جواب فصل 4 معادله ها و نامعادله ها ریاضی دهم

\((x + 1) (x – 3) = 0\)

الف

\(\left( {x{\rm{ }} + {\rm{ }}1} \right){\rm{ }}\left( {x{\rm{ }} - {\rm{ }}3} \right){\rm{ }} = {\rm{ }}0 \to x{\rm{ }} + {\rm{ }}1{\rm{ }} = {\rm{ }}0\) یا \(x{\rm{ }} - {\rm{ }}3{\rm{ }} = {\rm{ }}0\; \to \;x{\rm{ }} = {\rm{ }} - 1\) یا \(x = 3\)  

ب

پ

خیر؛ زیرا یکی از مقادیر منفی است و فقط عدد 3 جواب مسأله می باشد.

الف

\(\begin{array}{*{20}{l}}{{x^2} - 3x - 10 = 0}\\{\left( {x - 5} \right)\left( {x + 2} \right) = 0}\\{x - 5 = 0}\\{ \Rightarrow x = 5 \to {{\left( 5 \right)}^2} - 3\left( 5 \right) - 10 = 0}\\{x + 2 = 0}\\{ \Rightarrow x = - 2 \to {{\left( { - 2} \right)}^2} - 3\left( { - 2} \right) - 10 = 0}\end{array}\)

ب

\(\begin{array}{*{20}{l}}{t\left( {3t - 1} \right) = 0}\\{t = 0 \to \left( 0 \right)\left( {3 \times 0 - 1} \right) = 0}\\{3t - 1 = 0 \Rightarrow t = \frac{1}{3} \to \left( {\frac{1}{3}} \right)\left( {3 \times \frac{1}{3} - 1} \right) = 0}\end{array}\)

\(\begin{array}{*{20}{l}}{{x^2} = 25 \Rightarrow {x^2} - 25 = 0 \Rightarrow \left( {x - 5} \right)\left( {x + 5} \right) = 0}\\{x - 5 = 0 \Rightarrow x = 5}\\{x + 5 = 0 \Rightarrow x = - \:5}\end{array}\)

روش تجزیه به صورت زیر است:

\(\begin{array}{*{20}{l}}{{x^2} = 25 \Rightarrow {x^2} - 25 = 0 \Rightarrow \left( {x - 5} \right)\left( {x + 5} \right) = 0}\\{x - 5 = 0 \Rightarrow x = 5}\\{x + 5 = 0 \Rightarrow x = - \:5}\end{array}\)

روش ریشه گرفتن نیز به صورت زیر است:

\(x = \sqrt {25} \Rightarrow {x^2} = 25 \Rightarrow x = \pm 5\)

نتیجه می گیریم که از هر یک از روش های بالا برویم ، جواب بدست آمده صحیح می باشد.

خیر؛ این معادله فقط برای اعداد نامنفی یعنی \(a \ge 0\)   جواب دارد.

الف

\(x = \pm 10\,\,\,\,\, - \,\,\,\,\,\,{x^2} = 100\)

ب

\(x + 4 = 10 + 4 = 14\,\,\,\,\,\,\, - \,\,\,\,\,\,x = \pm 10\)

الف

\(\begin{array}{*{20}{l}}{{x^2} = 4}\\{x = 2}\\{x = - 2}\end{array}\)

ب

\(\begin{array}{*{20}{l}}{{t^2} = - 7}\\{t = \pm \sqrt { - 7} }\end{array}\)

جواب ندارد

پ

\(\begin{array}{*{20}{l}}{r - 2 = \pm 4}\\{r - 2 = 4 \to r = 6}\\{r - 2 = - 4 \to r = - 2}\end{array}\)

قطعه جدا شده در شکل 9 مربع دارد یعنی مربع عدد 3 ؛ در نتیجه عدد 3 را در پرانتز سمت راست تساوی و عدد 9 را در سمت چپ تساوی می گذاریم.

\({x^2} + 6x + 9 = {\left( {x + 3} \right)^2}\)

\({x^2} + ax + \frac{{{a^2}}}{4} = {(x + \frac{a}{2})^2}\)

    به دو طرف معادله 9 را اضافه کرده ایم تا سمت چپ مربع کامل شود \({x^2} - 6x + \;9 = - 4 + 9\)

الف

\(\begin{array}{*{20}{l}}{{x^2} + 2x + {{\left( {\frac{2}{2}} \right)}^2} = 24 + {{\left( {\frac{2}{2}} \right)}^2}}\\{{{\left( {x + 1} \right)}^2} = 25}\\{x + 1 = \pm 5}\\{x = 5 - 1 = 4}\\{x = - 5 - 1 = - 6}\end{array}\)

ب

\(\begin{array}{*{20}{l}}{{t^2} + 3t + {{\left( {\frac{3}{2}} \right)}^2} = 3 + {{\left( {\frac{3}{2}} \right)}^2}}\\{{{\left( {t + \frac{3}{2}} \right)}^2} = \frac{{45}}{4}}\\{t + \frac{3}{2} = \pm \frac{{3\sqrt 5 }}{2}}\\{t = - \frac{3}{2} + \frac{{3\sqrt 5 }}{2}}\\{t = - \frac{3}{2} - \frac{{3\sqrt 5 }}{2}}\end{array}\)

پ

\(\begin{array}{*{20}{l}}{{n^2} - 4n = - 5}\\{{n^2} - 4n + {{\left( {\frac{{ - 4}}{2}} \right)}^2} = - 5 + {{\left( {\frac{{ - 4}}{2}} \right)}^2}}\\{{{\left( {n - 2} \right)}^2} = - 1}\\{n - 2 = \pm \sqrt { - 1} \; \otimes }\end{array}\)

ت

\(\begin{array}{*{20}{l}}{2{\mkern 1mu} {r^2} + r = 2}\\{{r^2} + \frac{1}{2}r = 1}\\{{r^2} + \frac{1}{2}r + {{\left( {\frac{1}{4}} \right)}^2} = 1 + {{\left( {\frac{1}{4}} \right)}^2}}\\{{{\left( {r + \frac{1}{4}} \right)}^2} = \frac{{17}}{{16}}}\\{r + \frac{1}{4} = \pm \frac{{\sqrt {17} }}{4}}\\{r = - \frac{1}{4} + \frac{{\sqrt {17} }}{4}}\\{r = - \frac{1}{4} - \frac{{\sqrt {17} }}{4}}\end{array}\)

 به دو طرف معادله، \( - \frac{c}{a}\)  را اضافه کرده ایم  \({x^2} + \frac{b}{a}x = - \frac{c}{a}\)

  به دو طرف معادله، \(\frac{{{b^2}}}{{4{a^2}}}\)  را اضافه کرده ایم تا سمت چپ مربع کامل شود \({x^2} + \frac{b}{a}x + \frac{{{b^2}}}{{4a}} = - \frac{c}{a} + \frac{{{b^2}}}{{4a}}\)

الف

\(\begin{array}{*{20}{l}}\begin{array}{l}{x^2} - x + 1 = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 1}\\{b = - \,1}\\{c = 1}\end{array}} \right.\\\\ \Rightarrow \Delta = {b^2} - 4ac = {\left( { - \,1} \right)^2} - 4\left( 1 \right)\left( 1 \right) = - 3 < 0\end{array}\end{array}\)

جواب ندارد

ب

\(\begin{array}{*{20}{l}}\begin{array}{l} - 2{x^2} + x + 3 = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = - \,2}\\{b = \,1}\\{c = 3}\end{array}} \right.\\\\ \Rightarrow \Delta = {b^2} - 4ac = {\left( 1 \right)^2} - 4\left( { - 2} \right)\left( 3 \right) = 25\\\end{array}\\{x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - 1 \pm 5}}{{ - 4}} \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}{x_1} = \frac{3}{2}\\\end{array}\\{{x_2} = - 1}\end{array}} \right.}\end{array}\)

پ

\(\begin{array}{*{20}{l}}\begin{array}{l} - {x^2} + 4x - 4 = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = - 1}\\{b = \,4}\\{c = - 4}\end{array}} \right.\\\\ \Rightarrow \Delta = {b^2} - 4ac = {\left( 4 \right)^2} - 4\left( { - 1} \right)\left( { - 4} \right) = 0\\\end{array}\\{x = \frac{{ - b}}{{2a}} = \frac{{ - 4}}{{ - 2}} = 2}\end{array}\)

1

\(\begin{array}{l}{x^2} - 11x = - 10\\\\ \Rightarrow {x^2} - 11\,x + 10 = 0\\\\ \Rightarrow \left( {x - 10} \right)\left( {x - 1} \right) = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}x - 10 = 0 \Rightarrow x = 10\\\end{array}\\{x - 1 = 0 \Rightarrow x = 1}\end{array}} \right.\end{array}\)

2

\(\begin{array}{l}5{t^2} = 20\\\\ \Rightarrow {t^2} = 4 \Rightarrow {t^2} - 4 = 0\\\\ \Rightarrow \left( {t - 2} \right)\left( {t + 2} \right) = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}t - 2 = 0 \Rightarrow t = 2\\\end{array}\\{t + 2 = 0 \Rightarrow t = - 2}\end{array}} \right.\end{array}\)

3

\(\begin{array}{l}5{a^2} - 7a = 2a(a - 3)\\\\ \Rightarrow 5{a^2} - 7a = 2{a^2} - 6a\\\\ \Rightarrow 3{a^2} - a = 0\\\\ \Rightarrow a\left( {3a - 1} \right) = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}a = 0\\\end{array}\\{3a - 1 = 0 \Rightarrow a = \frac{1}{3}}\end{array}} \right.\end{array}\)

4

\(\begin{array}{l}4{k^2} - 12k + 8 = 0\\\\ \Rightarrow {k^2} - 3k + 2 = 0\\\\ \Rightarrow \left( {k - 2} \right)\left( {k - 1} \right) = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}k - 2 = 0 \Rightarrow k = 2\\\end{array}\\{k - 1 = 0 \Rightarrow k = 1}\end{array}} \right.\end{array}\)

1

\(\begin{array}{l}{n^2} - 2 = 26\\\\ \Rightarrow {n^2} = 28 \Rightarrow n = \pm \sqrt {28} = \pm 2\sqrt 7 \end{array}\)

2

\(\begin{array}{l}{x^2} + 12 = 3\\\\ \Rightarrow {x^2} = - 9 \Rightarrow x = \pm \sqrt { - 9} \otimes \end{array}\)

3

\(\begin{array}{l}{(3t - 2)^2} = 4\\\\\begin{array}{*{20}{l}}\begin{array}{l} \Rightarrow 3t - 2 = \pm \sqrt 4 = \pm 2\\\\ \Rightarrow 3t = 2 \pm 2 \Rightarrow t = \frac{{2 \pm 2}}{3}\\\end{array}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}t = \frac{{2 + 2}}{3} = \frac{4}{3}\\\end{array}\\{t = \frac{{2 - 2}}{3} = 0}\end{array}} \right.}\end{array}\end{array}\)

4

\(\begin{array}{l}3 - 3k = 3k(2k - 1)\\\\\begin{array}{*{20}{l}}\begin{array}{l} \Rightarrow 3 - 3k = 6{k^2} - 3k\\\\ \Rightarrow 3 = 6{k^2}\\\\ \Rightarrow \frac{3}{6} = {k^2} = \frac{1}{2}\\\end{array}\\{ \Rightarrow k = \pm \frac{{\sqrt 2 }}{2}}\end{array}\end{array}\)

1

\(\begin{array}{*{20}{l}}\begin{array}{l}{x^2} - 6x = 7\\\\ \Rightarrow {x^2} - 6x + {\left( {\frac{{ - 6}}{2}} \right)^2} = 7 + {\left( {\frac{{ - 6}}{2}} \right)^2}\\\\ \Rightarrow {\left( {x - 3} \right)^2} = 7 + 9 = 16\\\end{array}\\\begin{array}{l} \Rightarrow x - 3 = \pm 4\\\\ \Rightarrow x = 3 \pm 4 \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}x = 7\\\end{array}\\{x = - 1}\end{array}} \right.\end{array}\end{array}\)

2

\(\begin{array}{*{20}{l}}\begin{array}{l}{s^2} - 3x + 3 = 0\\\\ \Rightarrow {s^2} - 3s = - 3\\\\ \Rightarrow {s^2} - 3s + {\left( {\frac{{ - 3}}{2}} \right)^2} = - 3 + {\left( {\frac{{ - 3}}{2}} \right)^2} = - 0/75\\\end{array}\\\begin{array}{l} \Rightarrow {\left( {s - 2/25} \right)^2} = - 0/75\\\\ \Rightarrow s - 2/25 = \pm \sqrt { - 0/75} \otimes \end{array}\end{array}\)

3

\(\begin{array}{l}{r^2} + 4r + 4 = 0\\\\ \Rightarrow {\left( {r + 2} \right)^2} = 0 \Rightarrow r + 2 = 0 \Rightarrow r = - 2\end{array}\)

4

\(\begin{array}{*{20}{l}}\begin{array}{l}2{a^2} + 5a - 3 = 0\\\\ \Rightarrow 2{a^2} + 5a = 3\\\\ \Rightarrow {a^2} + \frac{5}{2}a = \frac{3}{2}\\\\ \Rightarrow {a^2} + \frac{5}{2}a + {\left( {\frac{5}{4}} \right)^2} = \frac{3}{2} + {\left( {\frac{5}{4}} \right)^2}\end{array}\\\begin{array}{l}\\ \Rightarrow {\left( {a + \frac{5}{4}} \right)^2} = \frac{{49}}{{16}}\\\\ \Rightarrow a + \frac{5}{4} = \pm \frac{7}{4}\\\\ \Rightarrow a = - \frac{5}{4} \pm \frac{7}{4}\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}a = - 3\\\end{array}\\{a = \frac{1}{2}}\end{array}} \right.\end{array}\end{array}\)

1

\(\begin{array}{*{20}{l}}\begin{array}{l}4{x^2} - 13x + 3 = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 4}\\{b = - 13}\\{c = 3}\end{array}} \right.\\\\ \Rightarrow \Delta = {b^2} - 4ac = {\left( { - 13} \right)^2} - 4\left( 4 \right)\left( 3 \right) = 121\\\end{array}\\\begin{array}{l} \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{13 \pm 11}}{{12}}\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}{x_1} = \frac{{13 + 11}}{8} = 3\\\end{array}\\{{x_2} = \frac{{13 - 11}}{8} = \frac{1}{4}}\end{array}} \right.\end{array}\end{array}\)

2

\(\begin{array}{*{20}{l}}\begin{array}{l}r - {r^2} = 3\\\\ \Rightarrow - {r^2} + r - 3 = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = - 1}\\{b = 1}\\{c = - 3}\end{array}} \right.\\\\ \Rightarrow \Delta = {b^2} - 4ac = {\left( 1 \right)^2} - 4\left( { - 1} \right)\left( { - 3} \right) = 1 - 12\\\end{array}\\{ \Rightarrow \Delta = - 11 < 0}\end{array}\)

جواب ندارد

3

\(\begin{array}{l}{a^2} + 2\sqrt 3 a = 9\\\\\begin{array}{*{20}{l}}\begin{array}{l} \Rightarrow {a^2} + 2\sqrt 3 a - 9 = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 1}\\{b = 2\sqrt 3 }\\{c = - 9}\end{array}} \right.\\\end{array}\\\begin{array}{l} \Rightarrow \Delta = {b^2} - 4ac = {\left( {2\sqrt 3 } \right)^2} - 4\left( 1 \right)\left( { - 9} \right)\\\end{array}\\\begin{array}{l} \Rightarrow \Delta = 48\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - 2\sqrt 3 \pm 4\sqrt 3 }}{2} = - \sqrt 3 \pm 2\sqrt 3 \\\end{array}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}{x_1} = - \,3\sqrt 3 \\\end{array}\\{{x_2} = \sqrt 3 }\end{array}} \right.}\end{array}\end{array}\)

4

\(\begin{array}{*{20}{l}}\begin{array}{l}\frac{{{t^2}}}{3} - \frac{t}{2} - \frac{3}{2} = 0\\\\\mathop \Rightarrow \limits^{ \times 6} \,2{t^2} - 3t - 4 = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 2}\\{b = - 3}\\{c = - 4}\end{array}} \right.\\\\ \Rightarrow \Delta = {b^2} - 4ac = {\left( { - 3} \right)^2} - 4\left( 2 \right)\left( { - 4} \right) = 41\\\end{array}\\\begin{array}{l} \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{3 \pm \sqrt {41} }}{4}\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}{x_1} = \frac{{3 + \sqrt {41} }}{4}\\\end{array}\\{{x_2} = \frac{{3 - \sqrt {41} }}{4}}\end{array}} \right.\end{array}\end{array}\)

1

با روش ریشه گیری:

\({x^2} = 125 \Rightarrow x = \pm 5\sqrt 5 \)

2

با روش مربع کامل:

\({\left( {z - 3} \right)^2} = 0 \Rightarrow z - 3 = 0 \Rightarrow z = 3\)

3

با روش فرمول کلی:

\(\begin{array}{*{20}{l}}\begin{array}{l}4{a^2} + 3a - 1 = 0 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{A = 4}\\{B = 3}\\{C = - 1}\end{array}} \right.\\\\ \Rightarrow \Delta = {B^2} - 4AC = {\left( 3 \right)^2} - 4\left( 4 \right)\left( { - 1} \right) = 25\\\end{array}\\\begin{array}{l} \Rightarrow a = \frac{{ - B \pm \sqrt \Delta }}{{2A}} = \frac{{ - 3 \pm \sqrt {25} }}{8}\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}{a_1} = \frac{{ - 3 + 5}}{8} = \frac{1}{4}\\\end{array}\\{{a_2} = \frac{{ - 3 - 5}}{8} = - 1}\end{array}} \right.\end{array}\end{array}\)

4

با روش فرمول کلی:

\(\begin{array}{*{20}{l}}\begin{array}{l} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{A = 1}\\{B = \sqrt 2 }\\{C = - 4}\end{array}} \right.\\\\ \Rightarrow \Delta = {B^2} - 4AC = {\left( {\sqrt 2 } \right)^2} - 4\left( 1 \right)\left( { - 4} \right) = 18\\\end{array}\\\begin{array}{l} \Rightarrow b = \frac{{ - B \pm \sqrt \Delta }}{{2A}} = \frac{{ - \sqrt 2 \pm 3\sqrt 2 }}{2}\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}{b_1} = \frac{{ - \sqrt 2 + 3\sqrt 2 }}{2} = \sqrt 2 \\\end{array}\\{{b_2} = \frac{{ - \sqrt 2 - 3\sqrt 2 }}{2} = - 2\sqrt 2 }\end{array}} \right.\end{array}\end{array}\)

فرض کنیم که n یک عدد فرد باشد؛ بنابراین 2n+ یک عدد فرد است. در نتیجه داریم :

\(\begin{array}{*{20}{l}}\begin{array}{l}{n^2} + {\left( {n + 2} \right)^2} = 290\\\\ \Rightarrow {n^2} + {n^2} + 4n + 4 = 2{n^2} + 4n + 4 = 290\\\end{array}\\\begin{array}{l} \Rightarrow {n^2} + 2n + 2 = 145\\\\ \Rightarrow {n^2} + 2n = 143\\\\ \Rightarrow {n^2} + 2n + {\left( {\frac{2}{2}} \right)^2} = 143 + {\left( {\frac{2}{2}} \right)^2}\\\end{array}\\\begin{array}{l} \Rightarrow {\left( {n + 1} \right)^2} = 144\\\\ \Rightarrow n + 1 = \pm 12\\\\ \Rightarrow n = - 1 \pm 12\\\end{array}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}n = - 1 + 12 \Rightarrow n = 11 \Rightarrow \begin{array}{*{20}{c}}{}&{}\end{array}11\,,\,13\\\end{array}\\{n = - 1 - 12 \Rightarrow n = - 13 \Rightarrow \begin{array}{*{20}{c}}{}&{}\end{array} - 13\,,\, - 11}\end{array}} \right.}\end{array}\)

\(\begin{array}{*{20}{l}}\begin{array}{l}a\,,\,4a + 3\\\\ \Rightarrow a\left( {4a + 3} \right) = 45\\\\ \Rightarrow 4{a^2} + 3a - 45 = 0 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{A = 4}\\{B = 3}\\{C = - 45}\end{array}} \right.\\\end{array}\\\begin{array}{l} \Rightarrow \Delta = {B^2} - 4AC = {\left( 3 \right)^2} - 4\left( 4 \right)\left( { - 45} \right) = 729\\\end{array}\\\begin{array}{l} \Rightarrow a = \frac{{ - B \pm \sqrt \Delta }}{{2A}} = \frac{{ - 3 \pm \sqrt {729} }}{8} = \frac{{ - 3 \pm 27}}{8}\\\end{array}\\\begin{array}{l} \Rightarrow {a_1} = \frac{{ - 3 - 27}}{8} = - \frac{{15}}{4}\, \otimes \\\end{array}\\{ \Rightarrow {a_2} = \frac{{ - 3 + 27}}{8} = 3}\\\begin{array}{l}\\ \Rightarrow a = 3\end{array}\end{array}\)

بنابراین:

عرض = 3  و  طول = 15

\(\begin{array}{*{20}{l}}\begin{array}{l}n\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} n + 4\,\,\,\\\\ \Rightarrow \left( {n + 4} \right)\left( {n + 4 + 4} \right) = 60\\\\ \Rightarrow \left( {n + 4} \right)\left( {n + 8} \right) = 60\\\end{array}\\\begin{array}{l} \Rightarrow {n^2} + 12n + 32 = 60\\\\ \Rightarrow {n^2} + 12n = 28\\\\ \Rightarrow {n^2} + 12n + {\left( {\frac{{12}}{2}} \right)^2} = 28 + {\left( {\frac{{12}}{2}} \right)^2}\\\end{array}\\\begin{array}{l} \Rightarrow {\left( {n + 6} \right)^2} = 64 \Rightarrow n + 6 = \pm \sqrt {64} \\\\ \Rightarrow n = - 6 \pm 8\\\end{array}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{n = 2}\\{n = - 14 \otimes }\end{array}} \right.\,\,\,\, \Rightarrow n = 2}\end{array}\)

2 سال و 6 سال سن دارند.

\(\begin{array}{*{20}{l}}\begin{array}{l}\left( {10 + 2x} \right)\left( {15 + 2x} \right) = 300\\\\ \Rightarrow 4{x^2} + 50x + 150 = 300\\\\ \Rightarrow 4{x^2} + 50x - 150 = 0\\\end{array}\\\begin{array}{l} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 4}\\{b = 50}\\{c = - 150}\end{array}} \right.\\\\ \Rightarrow \Delta = {b^2} - 4ac = {\left( {50} \right)^2} - 4\left( 4 \right)\left( { - 150} \right) = 4900\\\end{array}\\\begin{array}{l} \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - 50 \pm \sqrt {4900} }}{8} = \frac{{ - 50 \pm 70}}{8}\\\end{array}\\\begin{array}{l}{x_1} = \frac{{ - 50 - 70}}{8} = - 15\\\end{array}\\{{x_2} = \frac{{ - 50 + 70}}{8} = 2/5 \Rightarrow x = 2/5}\end{array}\)

بنابراین:

عرض = 15  و  طول = 20

اگر تعداد تیم ها را n  و هر تیم 1-n بازی دارد. پس با توجه به اینکه بین هر دو تیم تنها یک بازی صورت می گیرد و اصطلاحاً رفت و برگشتی نیست. در نتیجه کل بازی ها برابر \(\frac{{n\left( {n - 1} \right)}}{2}\)  است. بنابراین:

: الگو برای تعداد بازی ها \(\frac{{n\left( {n - 1} \right)}}{2}\) 

\(\begin{array}{*{20}{l}}\begin{array}{l}\frac{{n\left( {n - 1} \right)}}{2} = 45 \Rightarrow n\left( {n - 1} \right) = 90\\\\ \Rightarrow {n^2} - n - 90 = 0 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 1}\\{b = - 1}\\{c = - 90}\end{array}} \right.\\\end{array}\\\begin{array}{l} \Rightarrow \Delta = {b^2} - 4ac = {\left( { - 1} \right)^2} - 4\left( 1 \right)\left( { - 90} \right) = 361\\\end{array}\\\begin{array}{l}n = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{1 \pm 19}}{2} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{n_1} = \frac{{1 - 19}}{2} = - 9 \otimes }\\{{n_2} = \frac{{1 + 19}}{2} = 10}\end{array}} \right.\,\,\,\,\\\\ \Rightarrow n = 10\;\;\;\;N = n - 1\end{array}\end{array}\)

10 تیم

\(\begin{array}{*{20}{l}}\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}P = 0/006{s^2} - 0/02s + 120\\\end{array}\\{{P_1} = 125}\end{array}} \right.\\\\ \Rightarrow 125 = 0/006{s^2} - 0/02s + 120\\\\ \Rightarrow 0/006{s^2} - 0/02s - 5 = 0\\\end{array}\\\begin{array}{l} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 0/006}\\{b = - 0/02}\\{c = - 5}\end{array}} \right.\\\\ \Rightarrow \Delta = {b^2} - 4ac = {\left( { - 0/02} \right)^2} - 4\left( {0/006} \right)\left( { - 5} \right) = 0/1204\\\end{array}\\\begin{array}{l} \Rightarrow s = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{0/02 \pm \sqrt {0/1204} }}{{0/012}}\\\\ \Rightarrow s = \frac{{0/02 + \sqrt {0/1204} }}{{0/012}} \simeq 31\end{array}\end{array}\)

31 سال

الف

ب

نقطه (0, -4) پایین ترین نقطه این نمودار است؛ محور y ، محور تقارن این شکل است.

پ

با 5 نقطه ؛ بله ، با سه نقطه نیز می توان این شکل را رسم کرد.

ت

این منحنی محور x را در نقاط (-2, 0) و (2, 0) قطع می کند.

الف

\(y = {x^2} - 4x + 5\;\; \Rightarrow \;\;y = {\left( {x - 2} \right)^2} + 1\)

ب

\(x - 2 = 0\; \Rightarrow \;x = 2\)

پ

ت

طول این نقطه برابر با ریشه داخل پرانتز است و عرض آن از جایگذاری ریشه در تابع منحنی بدست می آید:

\(\begin{array}{*{20}{l}}{y = {{\left( {x - 2} \right)}^2} + 1 \Rightarrow x - 2 = 0 \Rightarrow {x_0} = 2 \Rightarrow {y_0} = {{\left( {2 - 2} \right)}^2} + 1 = 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = 2}\\{{y_0} = 1}\end{array}} \right. \Rightarrow \left( {2\,,\,1} \right)}\end{array}\)

الف

\(\begin{array}{*{20}{l}}\begin{array}{l}y = {\left( {x + 1} \right)^2} - 2 \Rightarrow x + 1 = 0 \Rightarrow {x_0} = - 1\\\\ \Rightarrow {y_0} = {\left( { - 1 + 1} \right)^2} - 2 = - 2\end{array}\\\begin{array}{l}\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = - 1}\\{{y_0} = - 2}\end{array}} \right. \Rightarrow \left( { - 1\,,\, - 2} \right)\end{array}\end{array}\)

نقطه رأس: (-1, -2)

\(\begin{array}{*{20}{l}}{{x_1} = - 2 \Rightarrow {y_1} = {{\left( { - 2 + 1} \right)}^2} - 2 = - 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 2}\\{{y_1} = - 1}\end{array}} \right. \Rightarrow \left( { - 2\,,\, - 1} \right)}\\{{x_2} = 0 \Rightarrow {y_2} = {{\left( {0 + 1} \right)}^2} - 2 = - 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = 0}\\{{y_2} = - 1}\end{array}} \right. \Rightarrow \left( {0\,,\, - 1} \right)}\end{array}\)

ب

\(\begin{array}{*{20}{l}}{y = - 2{x^2} + 1 \Rightarrow {x_0} = 0 \Rightarrow {y_0} = - 2{{\left( 0 \right)}^2} + 1 = 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = 0}\\{{y_0} = 1}\end{array}} \right. \Rightarrow \left( {0\,,\,1} \right)}\end{array}\)

نقطه رأس: (0, 1)

\(\begin{array}{*{20}{l}}{{x_1} = - 1 \Rightarrow {y_1} = - 2{{\left( { - 1} \right)}^2} + 1 = - 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 1}\\{{y_1} = - 1}\end{array}} \right. \Rightarrow \left( { - 1\,,\, - 1} \right)}\\{{x_2} = 1 \Rightarrow {y_2} = - 2{{\left( 1 \right)}^2} + 1 = - 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = 1}\\{{y_2} = - 1}\end{array}} \right. \Rightarrow \left( {1\,,\, - 1} \right)}\end{array}\)

الف

\(\begin{array}{*{20}{l}}\begin{array}{l}y = a{x^2} + bx + c\\\\ \Rightarrow y = a\left( {{x^2} + \frac{b}{a}x} \right) + c\\\\ \Rightarrow y = a\left( {{x^2} + \frac{b}{a}x + {{\left( {\frac{b}{{2a}}} \right)}^2} - {{\left( {\frac{b}{{2a}}} \right)}^2}} \right) + c\end{array}\\\begin{array}{l}\\ \Rightarrow y = y = a\left( {{x^2} + \frac{b}{a}x + \frac{{{b^2}}}{{4{a^2}}}} \right) + c - \frac{{{b^2}}}{{4a}}\\\\ \Rightarrow y = a{\left( {x + \frac{b}{{2a}}} \right)^2} + \frac{{4ac - {b^2}}}{{4a}}\end{array}\end{array}\)

ب

\(\begin{array}{*{20}{l}}\begin{array}{l}x + \frac{b}{{2a}} = 0\\\\ \Rightarrow {x_0} = - \frac{b}{{2a}}\\\\ \Rightarrow {y_0} = a{\left( { - \frac{b}{{2a}} + \frac{b}{{2a}}} \right)^2} + \frac{{4ac - {b^2}}}{{4a}} = \frac{{4ac - {b^2}}}{{4a}}\\\end{array}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}{x_0} = - \frac{b}{{2a}}\\\end{array}\\{{y_0} = \frac{{4ac - {b^2}}}{{4a}}}\end{array}} \right.}\end{array}\)

الف

ب

\(\begin{array}{*{20}{l}}{y = - {x^2} + 1}\\{ \Rightarrow {x_0} = 0 \Rightarrow {y_0} = - {{\left( 0 \right)}^2} + 1 = 0 + 1 = 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = 0}\\{{y_0} = 1}\end{array}} \right.}\\{ \Rightarrow {x_1} = - 1 \Rightarrow {y_1} = - {{\left( { - 1} \right)}^2} + 1 = - 1 + 1 = 0}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 1}\\{{y_1} = 0}\end{array}} \right.}\\{ \Rightarrow {x_2} = 1 \Rightarrow {y_2} = - {{\left( 1 \right)}^2} + 1 = - 1 + 1 = 0}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = - 1}\\{{y_2} = 0}\end{array}} \right.}\end{array}\)

الف

\(\begin{array}{*{20}{l}}{y = - {{\left( {x + 1} \right)}^2} - 3}\\{x + 1 = 0 \Rightarrow {x_0} = - 1 \Rightarrow {y_0} = - {{\left( { - 1 + 1} \right)}^2} - 3 = 0 - 3 = - 3}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = - 1}\\{{y_0} = - 3}\end{array}} \right.}\\{ \Rightarrow {x_1} = - 2 \Rightarrow {y_1} = - {{\left( { - 2 + 1} \right)}^2} - 3 = - 1 - 3 = - 4}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 2}\\{{y_1} = - 4}\end{array}} \right.}\\{ \Rightarrow {x_2} = 0 \Rightarrow {y_2} = - {{\left( {0 + 1} \right)}^2} - 3 = - 1 - 3 = - 4}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = 0}\\{{y_2} = - 4}\end{array}} \right.}\end{array}\)

ب

\(\begin{array}{*{20}{l}}{y = 3{x^2} - 2}\\{{x_0} = 0 \Rightarrow {y_0} = 3{{\left( 0 \right)}^2} - 2 = 0 - 2 = - 2}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = 0}\\{{y_0} = - 2}\end{array}} \right.}\\{{x_1} = - 1 \Rightarrow {y_1} = 3{{\left( { - 1} \right)}^2} - 2 = 3 - 2 = 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 1}\\{{y_1} = 1}\end{array}} \right.}\\{{x_2} = 1 \Rightarrow {y_2} = 3{{\left( 1 \right)}^2} - 2 = 3 - 2 = 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = 1}\\{{y_2} = 1}\end{array}} \right.}\end{array}\)

پ

\(\begin{array}{*{20}{l}}{y = x - {x^2} = - {{\left( {x - \frac{1}{2}} \right)}^2} + \frac{1}{4}}\\{x - \frac{1}{2} = 0 \Rightarrow {x_0} = \frac{1}{2} \Rightarrow {y_0} = - {{\left( {\frac{1}{2} - \frac{1}{2}} \right)}^2} + \frac{1}{4} = 0 + \frac{1}{4} = \frac{1}{4}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = \frac{1}{2}}\\{{y_0} = \frac{1}{4}}\end{array}} \right.}\\{{x_1} = - \frac{1}{2} \Rightarrow {y_1} = - {{\left( { - \frac{1}{2} - \frac{1}{2}} \right)}^2} + \frac{1}{4} = - 1 + \frac{1}{4} = - \frac{3}{4}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - \frac{1}{2}}\\{{y_1} = - \frac{3}{4}}\end{array}} \right.}\\{{x_2} = \frac{3}{2} \Rightarrow {y_2} = - {{\left( {\frac{3}{2} - \frac{1}{2}} \right)}^2} + \frac{1}{4} = - 1 + \frac{1}{4} = - \frac{3}{4}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = \frac{3}{2}}\\{{y_2} = - \frac{3}{4}}\end{array}} \right.}\end{array}\)

ت

\(\begin{array}{*{20}{l}}{y = \frac{{{x^2}}}{2} + x - 4 = \frac{1}{2}{{\left( {x + 1} \right)}^2} - \frac{9}{2}}\\{x + 1 = 0 \Rightarrow {x_0} = - 1 \Rightarrow {y_0} = \frac{1}{2}{{\left( { - 1 + 1} \right)}^2} - \frac{9}{2} = 0 - \frac{9}{2} = - \frac{9}{2}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = - 1}\\{{y_0} = - \frac{9}{2}}\end{array}} \right.}\\{{x_1} = - 2 \Rightarrow {y_1} = \frac{1}{2}{{\left( { - 2 + 1} \right)}^2} - \frac{9}{2} = \frac{1}{2} - \frac{9}{2} = - 4}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 2}\\{{y_1} = - 4}\end{array}} \right.}\\{{x_2} = 0 \Rightarrow {y_2} = \frac{1}{2}{{\left( {0 + 1} \right)}^2} - \frac{9}{2} = \frac{1}{2} - \frac{9}{2} = - 4}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = 0}\\{{y_2} = - 4}\end{array}} \right.}\end{array}\)

با توجه به اینکه عرض نقاط یکسان است، این دو نقطه نسبت به محور تقارن قرینه یکدیگرند؛ به عبارت دیگر محور تقارن از وسط طول های این دو نقطه می گذرد ؛ یعنی:

\({x_0} = \frac{{0 + ( - 2)}}{2} = - 1\)

در نتیجه خط 1-=x محور تقارن سهمی می باشد.

\(\begin{array}{l}\begin{array}{*{20}{l}}\begin{array}{l}\left( {0{\kern 1pt} ,{\kern 1pt} 2} \right) \Rightarrow 2 = a{\left( 0 \right)^2} + b\left( 0 \right) + c \Rightarrow c = 2\\\end{array}\\\begin{array}{l}\left( { - 1{\kern 1pt} ,{\kern 1pt} 0} \right) \Rightarrow 0 = a{\left( { - 1} \right)^2} + b\left( { - 1} \right) + 2 \Rightarrow a - b = - 2\\\end{array}\\\begin{array}{l}\left( {2{\kern 1pt} ,{\kern 1pt} 0} \right) \Rightarrow 0 = a{\left( 2 \right)^2} + b\left( 2 \right) + 2 \Rightarrow 4a + 2b = - 2\\\end{array}\\{\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}a - b = - 2\\\end{array}\\{4a + 2b = - 2}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}a = - 1\\\end{array}\\{b = - 1}\end{array}} \right. \Rightarrow y = - {x^2} + x + 2}\end{array}\\\\\begin{array}{*{20}{l}}{y = - {x^2} + x + 2 = - {{\left( {x - \frac{1}{2}} \right)}^2} + \frac{9}{4}}\\\begin{array}{l}\\x - \frac{1}{2} = 0 \Rightarrow {x_0} = \frac{1}{2} \Rightarrow {y_0} = - {\left( {\frac{1}{2} - \frac{1}{2}} \right)^2} + \frac{9}{4} = \frac{9}{4}\end{array}\\\begin{array}{l}\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = \frac{1}{2}}\\{{y_0} = \frac{9}{4}}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 1}\\{{y_1} = 0}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_2} = 2}\\{{y_2} = 0}\end{array}} \right.\end{array}\end{array}\end{array}\)

الف

مسیر پرتابه A:

\(\begin{array}{*{20}{l}}{y = - \frac{{{x^2}}}{2} + \frac{3}{2}x + 2 = - \frac{1}{2}{{\left( {x - \frac{3}{2}} \right)}^2} + \frac{{25}}{8}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = \frac{3}{2}}\\{{y_0} = \frac{{25}}{8}}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_1} = \frac{1}{2}}\\{{y_1} = \frac{{21}}{8}}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_2} = \frac{3}{2}}\\{{y_2} = \frac{{21}}{8}}\end{array}} \right.}\end{array}\)

مسیر پرتابه B:

\(\begin{array}{*{20}{l}}{y = - 2{x^2} + 3x + 2 = - 2{{\left( {x - \frac{3}{4}} \right)}^2} + \frac{{25}}{8}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = \frac{3}{4}}\\{{y_0} = \frac{{25}}{8}}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_1} = - \frac{1}{4}}\\{{y_1} = \frac{9}{8}}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_2} = \frac{7}{4}}\\{{y_2} = \frac{9}{8}}\end{array}} \right.}\end{array}\)

ب

طول های برخورد پرتابه A با زمین:

\(\begin{array}{*{20}{l}}{ - \frac{{{x^2}}}{2} + \frac{3}{2}x + 2 = - \frac{1}{2}{{\left( {x - \frac{3}{2}} \right)}^2} + \frac{{25}}{8} = 0}\\{ \Rightarrow \frac{1}{2}{{\left( {x - \frac{3}{2}} \right)}^2} = \frac{{25}}{8} \Rightarrow {{\left( {x - \frac{3}{2}} \right)}^2} = \frac{{25}}{4}}\\{ \Rightarrow x - \frac{3}{2} = \pm \frac{5}{2} \Rightarrow x = \frac{3}{2} \pm \frac{5}{2} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 1}\\{{x_2} = 4}\end{array}} \right.}\\{}\\{ \Rightarrow \Delta {x_A} = {x_2} - {x_1} = 4 + 1 = 5}\end{array}\)

طول های برخورد پرتابه B با زمین:

\(\begin{array}{*{20}{l}}{ - 2{x^2} + 3x + 2 = - 2{{\left( {x - \frac{3}{4}} \right)}^2} + \frac{{25}}{8} = 0}\\{ \Rightarrow 2{{\left( {x - \frac{3}{4}} \right)}^2} = \frac{{25}}{8} \Rightarrow {{\left( {x - \frac{3}{4}} \right)}^2} = \frac{{25}}{{16}}}\\{ \Rightarrow x - \frac{3}{4} = \pm \frac{5}{4} \Rightarrow x = \frac{3}{4} \pm \frac{5}{4} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - \frac{1}{4}}\\{{x_2} = 2}\end{array}} \right.}\\{ \Rightarrow \Delta {x_B} = {x_2} - {x_1} = 2 + \frac{1}{4} = \frac{9}{4}}\end{array}\)

پرتابه A مسیر افقی بیشتری رفته است:

\(\Delta {x_B} < \Delta {x_A}\)

پ

میزان ارتفاع همان عرض نقطه سهمی است که در این مورد هر دو عرض یکسان است، پس ارتفاع یکسانی پیموده اند؛ یعنی ارتفاع پیموده شده آنها \(\frac{{25}}{8}\)  است.

اگر ضریب x مثبت باشد، علامت قبل از ریشه منفی و بعد از آن مثبت است، اگر ضریب x منفی شود، علامت ها عکس خواهد شد، یعنی قبل از ریشه مثبت و بعد از آن منفی خواهد شد.

اگر a مثبت باشد، حالت نمودار به شکل زیر خواهد بود. با توجه به شکل قبل از ریشه منفی و بعد از ریشه مثبت است، یعنی قبل از ریشه، مخالف علامت a است و بعد از ریشه، موافق علامت a است؛ یعنی:

اگر a منفی باشد، حالت نمودار به شکل زیر خواهد بود. با توجه به شکل قبل از ریشه مثبت و بعد از ریشه منفی است، یعنی قبل از ریشه، مخالف علامت a است و بعد از ریشه، موافق علامت a است؛ یعنی:

\(x = - 1 \Rightarrow y = 5\left( { - 1} \right) - 2 = - 5 - 2 = - 7 \to \)منفی:

\(x = 3\,\,{\kern 1pt} \Rightarrow y = 5\left( 3 \right) - 2 = 15 - 2 = 13\;\;{\mkern 1mu} {\kern 1pt} \to \)مثبت:    

\(x = 0\,\,\, \Rightarrow \,\,y = \left( {2\left( 0 \right) - 1} \right)\left( {3 - \left( 0 \right)} \right) = \left( { - 1} \right)\left( 3 \right) = - 3 \to \)منفی:

\(x = 4 \Rightarrow y = \left( {2\left( 4 \right) - 1} \right)\left( {3 - \left( 4 \right)} \right) = \left( 7 \right)\left( { - 1} \right) = - 7 \to \)مثبت:   

الف

ب

پ

ت

\(\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{a > 0}\\{\Delta < 0}\end{array}} \right. \Rightarrow P\left( x \right) > 0}\\{}\\{\left\{ {\begin{array}{*{20}{l}}{a < 0}\\{\Delta < 0}\end{array}} \right. \Rightarrow P\left( x \right) < 0}\end{array}\)

\(\begin{array}{*{20}{l}}{y = - {x^2} + x + 2}\\\begin{array}{l} - {x^2} + x + 2 = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = - 1}\\{b = 1}\\{c = 2}\end{array}} \right. \Rightarrow \Delta = {b^2} - 4ac = {\left( 1 \right)^2} - 4\left( { - 1} \right)\left( 2 \right) = 9\\\end{array}\\\begin{array}{l}x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - 1 \pm 3}}{{ - 2}}\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}{x_1} = \frac{{ - 1 + 3}}{{ - 2}} = - 1\\\end{array}\\{{x_2} = \frac{{ - 1 - 3}}{{ - 2}} = 2}\end{array}} \right.\end{array}\end{array}\)

\(x^\circ = \frac{{ - 1 + 2}}{2} = \frac{1}{2} \Rightarrow y^\circ = \frac{9}{8}\,\, \Rightarrow \left( {\frac{1}{2}\,,\,\frac{9}{8}} \right)\)

الف

\(\left\{ {\begin{array}{*{20}{l}}{{x^2} - 9 = 0 \Rightarrow \left( {x - 3} \right)\left( {x + 3} \right) = 0 \Rightarrow x = - 3\,\,,\,\,x = 3}\\{3x - 1 = 0 \Rightarrow x = \frac{1}{3}}\end{array}} \right.\)

ب

\(\left\{ {\begin{array}{*{20}{l}}{ - {x^2} + 6x - 9 = 0 \Rightarrow - {{\left( {x - 3} \right)}^2} = 0 \Rightarrow x = 3}\\{{x^2} + x + 3 = 0 \Rightarrow \Delta = {{\left( 1 \right)}^2} - 4\left( 1 \right)\left( 3 \right) = - 11 < 0\,\, \otimes }\end{array}} \right.\)

الف

ب

\( - 1 < 3x \le 9\)  دو نامعادله را در \(\frac{1}{3}\)  ضرب می کنیم

\( - \frac{1}{3} < x \le 3\)

\(\begin{array}{*{20}{l}}\begin{array}{l}15 \le C \le 25 \Rightarrow 15 \le \frac{5}{9}\left( {F - 32} \right) \le 25\\\\ \Rightarrow \frac{9}{5} \times 15 \le F - 32 \le \frac{9}{5} \times 25\end{array}\\\begin{array}{l}\\ \Rightarrow 27 \le F - 32 \le 45\\\\ \Rightarrow 27 + 32 \le F \le 45 + 32 \Rightarrow 59 \le F \le 77\end{array}\end{array}\)

الف

\(-1 < x < 3\)

ب

\(y = {x^2} - 2x - 3 = \left( {x + 1} \right)\left( {x - 3} \right) \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x + 1 = 0 \Rightarrow x = - 1}\\{x - 3 = 0 \Rightarrow x = 3}\end{array}} \right.\)

پ

\({x^2} - 2x - 3 < 0\)  یعنی مجموعه مقادیری از x که به ازای آن ها 0>y باشد که با توجه به نمودار 

-1 < x < 3 مشخص شده است. همچنین به کمک جدول تعیین علامت نیز به همین جواب رسیدیم.

الف

\(\begin{array}{*{20}{l}}{{x^2} \le 4 \Rightarrow {x^2} - 4 \le 0}\\{{x^2} - 4 = 0 \Rightarrow {x^2} = 4 \Rightarrow x = \pm 2}\end{array}\)

ب

\(\begin{array}{*{20}{l}}{3{x^2} - x - 2 \ge 0}\\{3{x^2} - x - 2 = 0 \Rightarrow \Delta = {{\left( { - 1} \right)}^2} - 4\left( 3 \right)\left( { - 2} \right) = 25}\\{x = \frac{{ - \left( { - 1} \right) \pm \sqrt {25} }}{{2\left( 3 \right)}} = \frac{{1 \pm 5}}{6} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x = - \frac{2}{3}}\\{x = 1}\end{array}} \right.}\end{array}\)

\(x \in \left[ { - 3\,,\,3} \right]\)

\(x \in ( - \infty ,\; - 3]\; \cup \;[3\;,\; + \infty )\)

الف

\(\begin{array}{l}\left| {\frac{x}{3} + 1} \right| < \frac{2}{3} \Rightarrow \left| {\frac{{x + 3}}{3}} \right| < \frac{2}{3}\\\\ \Rightarrow \left| {x + 3} \right| < 2\\\\ \Rightarrow - 2 < x + 3 < 2 \Rightarrow - 5 < x < - 1\\\\ \Rightarrow x \in \left( { - 5\,,\, - 1} \right)\end{array}\)

ب

\(\begin{array}{l}\left| {5 - 2x} \right| \ge 1 \Rightarrow \left| {2x - 5} \right| \ge 1\\\\ \Rightarrow - 1 \le 2x - 5 \le 1 \Rightarrow 4 \le 2x \le 6\\\\ \Rightarrow 2 \le x \le 3 \Rightarrow x \in \left[ {2\,,\,3} \right]\end{array}\)

\(\begin{array}{*{20}{l}}{x \in \left( {a\,,\,b} \right) \Leftrightarrow \left| {x - r} \right| < d \Rightarrow \left\{ {\begin{array}{*{20}{l}}{r = \frac{{a + b}}{2}}\\{d = \frac{{b - a}}{2}}\end{array}} \right.}\\{ \Rightarrow x \in \left( {1\,,\,9} \right) \Rightarrow \left\{ {\begin{array}{*{20}{l}}{r = \frac{{9 + 1}}{2} = \frac{{10}}{2} = 5}\\{d = \frac{{9 - 1}}{2} = \frac{8}{2} = 4}\end{array}} \right. \Rightarrow \left| {x - 5} \right| < 4}\end{array}\)

\(\begin{array}{*{20}{l}}{x \in \left( {\left. { - \infty \,,\,a} \right]} \right. \cup \left[ {\left. {b\,,\,\infty } \right)} \right. \Leftrightarrow \left| {x - r} \right| > d \Rightarrow \left\{ {\begin{array}{*{20}{l}}{r = \frac{{a + b}}{2}}\\{d = \frac{{b - a}}{2}}\end{array}} \right.}\\{ \Rightarrow x \in \left( {\left. { - \infty \,,\,3} \right]} \right. \cup \left[ {\left. {6\,,\,\infty } \right)} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}{r = \frac{{6 + 3}}{2} = \frac{9}{2}}\\{d = \frac{{6 - 3}}{2} = \frac{3}{2}}\end{array}} \right. \Rightarrow \left| {x - \frac{9}{2}} \right| > \frac{3}{2}}\end{array}\)

الف

\(1 < 2x - 3 \le 3 \Rightarrow 4 < 2x \le 6 \Rightarrow 2 < x \le 3 \Rightarrow x \in \left( {\left. {2\,,\,3} \right]} \right.\)

ب

\(\begin{array}{*{20}{l}}\begin{array}{l}x + 1 \le 5 - x < 2x + 3\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}x + 1 \le 5 - x \Rightarrow 2x \le 4 \Rightarrow x \le 2 \Rightarrow x \in \left( {\left. { - \infty \,,\,2} \right]} \right.\\\end{array}\\{5 - x < 2x + 3 \Rightarrow 2 < 3x \Rightarrow \frac{2}{3} < x \Rightarrow x \in \left( {\frac{2}{3}\,,\,\infty } \right)}\end{array}} \right.\end{array}\\{}\\{ \Rightarrow x \in \left( {\left. { - \infty \,,\,2} \right]} \right. \cap \left( {\frac{2}{3}\,,\,\infty } \right) \Rightarrow x \in \left( {\left. {\frac{2}{3}\,,\,2} \right]} \right.}\end{array}\)

پ

\(\begin{array}{l} - 2 < \frac{{5 - x}}{2} < 0\,\mathop {\, \Rightarrow }\limits^{ \times \left( { - 1} \right)} \,\,0 < \frac{{x - 5}}{2} < 2\\\\ \Rightarrow 0 < x - 5 < 4 \Rightarrow 5 < x < 9 \Rightarrow x \in \left( {5\,,\,9} \right)\end{array}\)

ت

\(A = \frac{{4 - 2x}}{{3x + 1}} \ge 0\,\,\, \Rightarrow \left\{ {\begin{array}{*{20}{l}}{4 - 2x = 0 \Rightarrow x = 2}\\{3x + 1 = 0 \Rightarrow x = - \frac{1}{3}}\end{array}} \right.\)

\(x \in ( - \infty \,,\, - \frac{1}{3}) \cup \left[ {\left. {2\,,\,\infty } \right)} \right.\)

ث

\(B = x\left( {{x^2} + 4} \right) < 0\,\, \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x = 0}\\{{x^2} + 4 = 0 \Rightarrow \Delta = {{\left( 0 \right)}^2} - 4\left( 1 \right)\left( 4 \right) = - 16 < 0}\end{array}} \right.\)

\(x \in \left( { - \infty {\kern 1pt} ,\,0} \right)\)

ج

\(\begin{array}{*{20}{l}}{C = \frac{{{x^3} - x}}{{{x^2} - 2x + 2}} = \frac{{x\left( {{x^2} - 1} \right)}}{{{x^2} - 2x + 2}} \le 0}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x = 0}\\{{x^2} - 1 = 0 \Rightarrow {x^2} = 1 \Rightarrow x = \pm 1}\\{{x^2} - 2x + 2 = 0 \Rightarrow \Delta = {{\left( { - 2} \right)}^2} - 4\left( 1 \right)\left( 2 \right) = - 4 < 0}\end{array}} \right.}\end{array}\)

\(x \in \left( {\left. { - \infty {\kern 1pt} ,\, - 1} \right]} \right. \cup \left[ {0\,,\,1} \right]\)

چ

\(\begin{array}{l}\left| {7 - 2x} \right| < 1 \Rightarrow \left| {2x - 7} \right| < 1 \Rightarrow - 1 < 2x - 7 < 1\\\\ \Rightarrow 6 < 2x < 8 \Rightarrow 3 < x < 4 \Rightarrow x \in \left( {3\,,\,4} \right)\end{array}\)

ح

\(\begin{array}{*{20}{l}}\begin{array}{l}\left| {\frac{{x - 1}}{2} - 1} \right| \ge 3 \Rightarrow \left| {\frac{{x - 3}}{2}} \right| \ge 3 \Rightarrow \left| {x - 3} \right| \ge 6\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}x - 3 > 6 \Rightarrow x > 9 \Rightarrow x \in \left( {9\,,\,\infty } \right)\\\end{array}\\{x - 3 < - 6 \Rightarrow x < - 3 \Rightarrow x \in \left( { - \infty \,,\, - 3} \right)}\end{array}} \right.\\\end{array}\\{x \in \left( { - \infty \,,\, - 3} \right) \cup \left( {9\,,\,\infty } \right)}\end{array}\)

\(\begin{array}{*{20}{l}}{A = {x^2} + 3x + k}\\{\forall x \in \mathbb{R}:A > 0 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a = 1 > 0}\\{\Delta < 0}\end{array}} \right.}\\{\Delta < 0 \Rightarrow {b^2} - 4ac > 0 \Rightarrow {{\left( 3 \right)}^2} - 4\left( 1 \right)\left( k \right) < 0}\\{ \Rightarrow 9 - 4k < 0 \Rightarrow k > \frac{9}{4}}\end{array}\)

\(\begin{array}{l}y = m{x^2} - mx - 1\\\\\forall x \in \mathbb{R}:y < 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}a < 0 \Rightarrow m < 0\;\;{\mkern 1mu} {\kern 1pt} \left( 1 \right)\\\end{array}\\{\Delta < 0 \Rightarrow {b^2} - 4ac < 0 \Rightarrow {m^2} + 4m < 0\;\;{\mkern 1mu} {\kern 1pt} \left( 2 \right)}\end{array}} \right.\\\\\left( 2 \right):\\\\{m^2} + 4m < 0 \Rightarrow {m^2} + 4m = 0\\\\ \Rightarrow m\left( {m + 4} \right) = 0 \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}m = 0\\\end{array}\\{m = - 4}\end{array}} \right.\end{array}\)

\(\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{h = - 5{t^2} + 18t + 13}\\{h > 13 \Rightarrow t \in \left( {{t_1}\,,\,{t_2}} \right)}\end{array}} \right. \Rightarrow {t_1},{t_2} = ?}\\{ - 5{t^2} + 18t + 13 > 13 \Rightarrow - 5{t^2} + 18t > 0}\\{ \Rightarrow - 5{t^2} + 18t = 0 \Rightarrow t\left( { - 5t + 18} \right) = 0 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{t_1} = 0}\\{ - 5t + 18 = 0 \Rightarrow {t_2} = \frac{{18}}{5}}\end{array}} \right.}\end{array}\)

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}y = \frac{{15}}{8}{x^2} - 30x + 200\\\end{array}\\{y > 110 \Rightarrow x \in ?}\end{array}} \right.\\\\\frac{{15}}{8}{x^2} - 30x + 200 > 110\\\\ \Rightarrow \frac{{15}}{8}{x^2} - 30x + 90 > 0\\\\\frac{{15}}{8}{x^2} - 30x + 90 = 0\\\\ \Rightarrow \Delta = {\left( { - 30} \right)^2} - 4(\frac{{15}}{2})\left( {90} \right) = 900 - 675 = 225 > 0\\\\x = \frac{{ - \left( { - 30} \right) \pm \sqrt {225} }}{{2(\frac{{15}}{8})}} = \frac{{30 \pm 15}}{{\frac{{15}}{4}}} = \\\\\frac{{120 \pm 60}}{{15}} = 8 \pm 4\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}x = 12\\\end{array}\\{x = 4}\end{array}} \right.\end{array}\)