نمونه سوالات فصل 2 ریاضی نهم همراه با جواب تشریحی

نمایش به صورت مجموعه با اعضا:

\([ + 7,\infty )\)

نمایش به صورت نماد های ریاضی:

\(\{ x|x \in \mathbb{R},x \ge 7\} \)

نمایش روی محور:

\(\begin{array}{l}1) - \frac{2}{7} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{3}{5} \Rightarrow [7,5] = 35\\\\ \Rightarrow \frac{{ - 10}}{{35}} < \frac{1}{{35}} < \frac{2}{{35}} < \frac{3}{{35}} < \frac{4}{{35}} < \frac{{21}}{{35}} = \\\\ \Rightarrow \frac{{ - 2}}{7} < \frac{1}{{35}} < \frac{2}{{35}} < \frac{3}{{35}} < \frac{4}{{35}} < \frac{3}{5}\end{array}\)

\(\begin{array}{l}2)\frac{1}{8} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{1}{3} \Rightarrow [8,3] = 24\\\\ \Rightarrow \frac{3}{{24}} < \frac{4}{{24}} < \frac{5}{{24}} < \frac{6}{{24}} < \frac{7}{{24}} < \frac{8}{{24}} = \\\\ \Rightarrow \frac{1}{8} < \frac{1}{6} < \frac{5}{{24}} < \frac{1}{4} < \frac{7}{{24}} < \frac{1}{3}\end{array}\)

اعداد گویا عبارتند از:

\(\frac{{\sqrt {343} }}{{\sqrt 7 }},0,\frac{{1 + 3 \times 4}}{{8 - 2}},\frac{{{4^2} - 7 \times 8 + 1}}{{\sqrt {25} - \sqrt {16} }}\)

\(\frac{{\sqrt 4 }}{{\sqrt {25} }} < \frac{5}{9} < \frac{5}{8} < \frac{3}{4} < \frac{8}{5}\)

\(\begin{array}{l}1)3/4678678678... = 3/4\overline {678} \\\\2)6/023023023... = 6/\overline {023} \\\\3)81/111... = 81/\overline {01} \end{array}\)

\(\begin{array}{l}1)1/\overline {43} = 1/434343...\,\,,\,\,1/4\overline 3 = 1/4333...\\\\ \Rightarrow 1/\overline {43} > 1/4\overline 3 \\\\2)8/0\overline {101} = 8/0101101101...\,\,,\,\,8/\overline {01} = 8/010101...\\\\ \Rightarrow 8/0\overline {101} > 8/\overline {01} \end{array}\)

مختوم\(1)\frac{6}{8} = \frac{3}{4} = 0/75 \Rightarrow \)

متناوب ساده\(2)\frac{{19}}{{57}} = \frac{1}{3} = 0/\overline 3 \Rightarrow \)

متناوب مرکب\(3)\frac{{130}}{{1200}} = \frac{{13}}{{120}} = \frac{{13}}{{12 \times 2 \times 5}} \Rightarrow \)

متناوب ساده\(4)\frac{{24}}{{26}} = \frac{{12}}{{13}} \Rightarrow \)

متناوب مرکب\(5)\frac{{102}}{{36}} = \frac{{51}}{{18}} = \frac{{17}}{6} \Rightarrow \)

\(\begin{array}{l}1)7/1\overline 4 + 1/\overline 3 = 7/1\overline 4 + 1/3\overline 3 = 7/4\overline 7 \\\\2)8/\overline 8 - 3/\overline {35} = 8/\overline {88} - 3/\overline {35} = 5/\overline {53} \end{array}\)

\(\frac{1}{{1024}},\frac{5}{2},\frac{1}{{40}}\)

\(\frac{2}{{11}},\frac{1}{3},\frac{1}{7}\)

\(\frac{2}{{11}} = 0/181818 = 0/\overline {18} \)

رقم هفتم برابر 1 و رقم بیستم برابر 8 می باشد.

\(\begin{array}{l}2\frac{1}{3} - 3\frac{1}{2} \div (4\frac{1}{5} \div \frac{3}{4}) = \\\\ \Rightarrow \frac{7}{3} - \frac{7}{2} \div (\frac{{21}}{5} \div \frac{3}{4}) = \frac{7}{3} - \frac{7}{2} \div (\frac{{21}}{5} \times \frac{4}{3}) = \\\\ \Rightarrow \frac{7}{3} - \frac{7}{2} \div (\frac{{28}}{5}) = \frac{7}{3} - \frac{7}{2} \times \frac{5}{{28}} = \frac{7}{3} - \frac{5}{8} = \\\\ \Rightarrow \frac{{56 - 15}}{{24}} = \frac{{41}}{{24}} = 1\frac{{17}}{{24}}\end{array}\)

\(\begin{array}{l}\frac{{( - \frac{{24}}{{10}} + \frac{5}{4}) + (\frac{9}{8} + \frac{{ - 12}}{5})}}{{2(\frac{3}{5} + \frac{9}{{40}}) - 3(\frac{1}{5} + \frac{2}{3})}} = \\\\ \Rightarrow \frac{{(\frac{{ - 96 + 50}}{{40}}) + (\frac{{45 - 96}}{{40}})}}{{2(\frac{{24 + 9}}{{40}}) - 3(\frac{{3 + 10}}{{15}})}} = \\\\ \Rightarrow \frac{{\frac{{ - 96 + 50 + 45 - 96}}{{40}}}}{{\frac{{24 + 9}}{{20}} - \frac{{13}}{5}}} = \\\\ \Rightarrow \frac{{ - \frac{{97}}{{40}}}}{{\frac{{33 - 52}}{{20}}}} = \frac{{\frac{{ - 97}}{{40}}}}{{ - \frac{{19}}{{20}}}} = \frac{{ - 97 \times 20}}{{ - 19 \times 40}} = \frac{{97}}{{38}}\end{array}\)

\(\begin{array}{l}\frac{{( - \frac{{24}}{{10}} + \frac{5}{4}) + (\frac{9}{8} + \frac{{ - 12}}{5})}}{{2(\frac{3}{5} + \frac{9}{{40}}) - 3(\frac{1}{5} + \frac{2}{3})}} = \\\\ \Rightarrow \frac{{(\frac{{ - 96 + 50}}{{40}}) + (\frac{{45 - 96}}{{40}})}}{{2(\frac{{24 + 9}}{{40}}) - 3(\frac{{3 + 10}}{{15}})}} = \\\\ \Rightarrow \frac{{\frac{{ - 96 + 50 + 45 - 96}}{{40}}}}{{\frac{{24 + 9}}{{20}} - \frac{{13}}{5}}} = \\\\ \Rightarrow \frac{{ - \frac{{97}}{{40}}}}{{\frac{{33 - 52}}{{20}}}} = \frac{{\frac{{ - 97}}{{40}}}}{{ - \frac{{19}}{{20}}}} = \frac{{ - 97 \times 20}}{{ - 19 \times 40}} = \frac{{97}}{{38}}\end{array}\)

\(\begin{array}{l}2 + \frac{{2 + \frac{{2 + \frac{1}{2}}}{2}}}{{2 - \frac{{2 - \frac{1}{2}}}{2}}} = 2 + \frac{{2 + \frac{{\frac{5}{2}}}{2}}}{{2 - \frac{{\frac{3}{2}}}{2}}} = 2 + \frac{{2 + \frac{5}{4}}}{{2 - \frac{3}{4}}} = \\\\ \Rightarrow 2 + \frac{{\frac{{13}}{4}}}{{\frac{5}{4}}} = 2 + \frac{{13}}{5} = \frac{{10 + 13}}{5} = \frac{{23}}{5}\end{array}\)

\(\begin{array}{l}\frac{y}{x} = \frac{{80}}{{48}} = \frac{5}{3} \Rightarrow 3y = 5x\\\\y = x + 250 \Rightarrow \times 3 \Rightarrow 3y = 3x + 750\\\\ \Rightarrow 5x = 3x + 750 \Rightarrow 5x - 3x = 750\\\\ \Rightarrow 2x = 750 \Rightarrow x = \frac{{750}}{2} = 375\\\\ \Rightarrow y = x + 250 \Rightarrow y = 375 + 250 = \\\\ \Rightarrow y = 625\end{array}\)

الف)

\(4 < \sqrt {17} < 5 \Rightarrow - 5 < - \sqrt {17} < - 4\)

ب)

\(\begin{array}{l}4 < \sqrt {18} < 5 \Rightarrow 5 + 4 < 5 + \sqrt {18} < 5 + 5\\\\ \Rightarrow 9 < 5 + \sqrt {18} < 10\end{array}\)

)پ

\(\begin{array}{l}5 < \sqrt {28} < 6 \Rightarrow - 6 < - \sqrt {28} < - 5\\\\ \Rightarrow 7 - 6 < 7 - \sqrt {28} < 7 - 5\\\\ \Rightarrow 1 < 7 - \sqrt {28} < 2\end{array}\)

ت)

\(\begin{array}{l}2\sqrt 5 = \sqrt {20} \Rightarrow 4 < \sqrt {20} < 5\\\\ \Rightarrow 4 < 2\sqrt 5 < 5 \Rightarrow - 5 < - 2\sqrt 5 < - 4\\\\ \Rightarrow 3 - 5 < 3 - 2\sqrt 5 < 3 - 4\\\\ \Rightarrow - 2 < 3 - 2\sqrt 5 < - 1\end{array}\)

\(A = - 2 - \sqrt 6 \)

\(\begin{array}{l}1)|5\sqrt 5 - 7\sqrt 5 | - |7\sqrt 6 - 6\sqrt 7 | = \\\\ \Rightarrow - (5\sqrt 7 - 7\sqrt 5 ) - (7\sqrt 6 - 6\sqrt 7 ) = \\\\ \Rightarrow - 5\sqrt 7 + 7\sqrt 5 - 7\sqrt 6 + 6\sqrt 7 = \\\\ \Rightarrow \sqrt 7 + 7(\sqrt 5 - \sqrt 6 ) = \sqrt 7 - 7(\sqrt 6 - \sqrt 5 )\\\\2)|10 - 4\sqrt 5 | - |4\sqrt 5 - 9| = \\\\ \Rightarrow (10 - 4\sqrt 5 ) + (9 - 4\sqrt 5 ) = 19 - 8\sqrt 5 \end{array}\)

\(\begin{array}{l}1)a < 0,\frac{{ - \sqrt {{a^2}} }}{{ - |a|}} = \frac{{ - |a|}}{{ - |a|}} = \frac{{|a|}}{{|a|}} = 1\\\\2)a > 0,\frac{{|a|}}{{ - a}} = \frac{a}{{ - a}} = - 1\end{array}\)

\(\begin{array}{l}x < - 3 \Rightarrow x + 3 < 0\\\\1)\sqrt {{{(x + 3)}^2}} + \sqrt {{{( - 3 - x)}^2}} = \\\\ \Rightarrow |x + 3| + | - 3 - x| = |x + 3| + |x + 3| = \\\\ \Rightarrow 2|x + 3| = - 2(x + 3) = - 2x - 6\\\\2)\sqrt {{{(|x + 3| + | - 3 - x|)}^2}} = \\\\ \Rightarrow \sqrt {{{(|x + 3| + |x + 3|)}^2}} = \sqrt {{{(2|x + 3|)}^2}} = \\\\ \Rightarrow |2|x + 3|| = 2|x + 3| = \\\\ \Rightarrow - 2(x + 3) = - 2x - 6\end{array}\)

\(\frac{{11}}{{112}}\)

\(2/\overline 3 \)

2 و 5

گنگ

مجموعه اعداد گویا

\(\frac{{136}}{{99}}\)

گنگ

1 (یک)

5-

گنگ

6- و 5-

گویا

6

گویا

مثبت

\( - 7 - \sqrt {11} + 3\sqrt 2 \)

منفی

b-a

مبدأ

اجتماع

هر کسری که صورت و مخرجش عدد صحیح بوده و مخرجش غیر صفر باشد، عددی گویاست.

مجموعه اعداد گویا را نمی توان با اعضایش نمایش داد.

\(\frac{8}{3}\) یک کسر متناوب می باشد و نامتناهی.

بین دو عدد 0 و \(\sqrt 2 \) ، بیشمار عدد گویا وجود دارد.

مثال نقض:\(| - 3| > |0|\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\, - 3 < 0\)

\(\frac{{16}}{5} = 3/2\)

\(\begin{array}{l}\frac{{a + \frac{1}{{a + \frac{1}{a}}}}}{a} = \frac{a}{a} + \frac{{\frac{1}{{a + \frac{1}{a}}}}}{a} = 1 + \frac{1}{{a(a + \frac{1}{a})}} = \\\\ \Rightarrow 1 + \frac{1}{{{a^2} + 1}} = 1 + \frac{1}{{{{(\frac{1}{3})}^2} + 1}} = 1 + \frac{1}{{\frac{1}{9} + 1}} = \\\\ \Rightarrow 1 + \frac{1}{{\frac{{10}}{9}}} = 1 + \frac{9}{{10}} = 1 + 0/9 = 1/9\end{array}\)

\(\begin{array}{l}x = 0/\overline 9 \Rightarrow 10x = 9/\overline 9 \Rightarrow 10x - x = \\\\9/\overline 9 - 0/\overline 9 = 9 \Rightarrow 9x = 9 \Rightarrow x = 1\end{array}\)

بررسی گزینه ها:

الف) نادرست؛ مثال نقض: \(\sqrt 2 \times \sqrt 3 = \sqrt 6 \)

ب) نادرست؛ مثال نقض: \(\sqrt 2 + \sqrt 3 \)  که عددی گویا نیست.

پ) درست

\(\begin{array}{l}x = 2/\overline 7 \Rightarrow 10x = 27/\overline 7 \Rightarrow 10x - x = \\\\ \Rightarrow 27/\overline 7 - 2/\overline 7 = 25 \Rightarrow 9x = 25\\\\ \Rightarrow x = \frac{{25}}{9} \Rightarrow {(\frac{a}{b})^2} = \frac{{25}}{9} \Rightarrow \frac{a}{b} = \sqrt {\frac{{25}}{9}} = \frac{5}{3}\\\\ \Rightarrow \left\{ \begin{array}{l}a = 5\\b = 3\end{array} \right. \Rightarrow a + 2b = 11\end{array}\)

\(\begin{array}{l}\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \cdots + \frac{{n - 1}}{n} = x\\\\ \Rightarrow \times {n^2}\\\\ \Rightarrow \frac{{{n^2}}}{2} + \frac{{2{n^2}}}{3} + \frac{{3{n^2}}}{4} + \cdots + \frac{{{n^2}(n - 1)}}{n} = {n^2}x\end{array}\)

\(\begin{array}{l}\sqrt {16} < \sqrt {24} < \sqrt {25} \Rightarrow 4 < \sqrt {24} < 5 \Rightarrow \\\\ \Rightarrow - 5 < - \sqrt {24} < - 4 \Rightarrow 0 < 5 - \sqrt {24} < 1\end{array}\)

\(A = \{ x|x \in \mathbb{R}, - 2 \le x < 2\} = [ - 2,2)\)

\(\begin{array}{l}\left. \begin{array}{l}x = 0/\overline {123} \\\\1000x = 123/\overline {123} \end{array} \right\} \Rightarrow 1000x - x = 123\\\\ \Rightarrow 999x = 123 \Rightarrow x = \frac{{123}}{{999}} = \frac{{41}}{{333}}\end{array}\)

\(\begin{array}{l}x = 0/\overline 3 \Rightarrow 10x = 3/\overline 3 \\\\ \Rightarrow 10x - x = 3/\overline 3 - 0/\overline 3 = 3 \Rightarrow 9x = 3\\\\ \Rightarrow x = \frac{3}{9} \Rightarrow x = \frac{1}{3} \Rightarrow 0/\overline 3 + {(0/\overline 3 )^3} = \\\\\frac{1}{3} + {(\frac{1}{3})^3} = \frac{1}{3} + \frac{1}{{27}} = \frac{{9 + 1}}{{27}} = \frac{{10}}{{27}} = 0/370 \cdots \\\\ = 0/\overline {370} \end{array}\)

\(\left. \begin{array}{l}a > 0 \Rightarrow |a| = a\\\\b < 0 \Rightarrow |a| = - b\end{array} \right\} \Rightarrow |a| + |b| = a - b\)

\(\begin{array}{l}|5x - 2| + |3x - 1| + 4 - 10x = \\\\ \Rightarrow | - 27| + | - 16| + 4 + 50 = \\\\ \Rightarrow 27 + 16 + 4 + 50 = 97\end{array}\)

\(\begin{array}{l}a < b \Rightarrow a - b < 0 \Rightarrow |a - b| = \\\\ \Rightarrow - (a - b) = b - a\\\\a < 0 \Rightarrow - a > 0 \Rightarrow 3 - a > 0 = \\\\ \Rightarrow |3 - a| = 3 - a\\\\b > 0 \Rightarrow b + 1 > 0 \Rightarrow |b + 1| = b + 1\\\\ \Rightarrow |a - b| + |3 - a| + |b + 1| = \\\\ \Rightarrow b - a + 3 - a + b + 1 = - 2a + 2b + 4\end{array}\)

\(\begin{array}{l}|b| > |a| \Rightarrow b + a > 0 \Rightarrow |a + b| + |a| + |b| = \\\\ \Rightarrow a + b - a + b = 2b\end{array}\)

وقتی قدر مطلق برابر منفی عبارت داخل آن شده، یعنی \({m^n}\)  منفی بوده است؛ بنابراین n عددی فرد و m عددی منفی می باشد.

وقتی فاصله دو عدد x و y روی محور اعداد برابر \(\sqrt 3 \)  می باشد، یعنی:

\(\begin{array}{l}|x - y| = |y - x| = \sqrt 3 \\\\| - 3(x - y)| = | - 3||x - y| = 3\sqrt 3 \end{array}\)

برای گزینه 4 ، مثال نقض می آوریم:

\(\begin{array}{l}\left\{ \begin{array}{l}x = 1 - \sqrt 3 \\\\y = 1 + \sqrt 3 \end{array} \right.\left\{ \begin{array}{l}\sqrt {{x^2}} - \sqrt {{y^2}} = |x| - |y| = \\\\|x - y| = |(1 - \sqrt 3 ) - (1 + \sqrt 3 )|\end{array} \right.\\\\ \Rightarrow \left\{ \begin{array}{l}|1 - \sqrt 3 | - |1 + \sqrt 3 | = (\sqrt 3 - 1) - (\sqrt 3 + 1) = - 2\\\\|2| = 2\end{array} \right.\\\\ \Rightarrow - 2 \ne 2\end{array}\)

\(\begin{array}{l}x < 2 \Rightarrow 2x < 4 \Rightarrow 2x - 4 < 0 \Rightarrow |2x - 4 = 4 - 2x\\\\x < 2 \Rightarrow 3x < 6 = 3x - 6 < 0 \Rightarrow |3x - 6| = 6 - 3x\\\\x < 2 \Rightarrow 5x < 10 \Rightarrow 5x < 16 \Rightarrow 16 - 5x > 0\\\\ \Rightarrow |16 - 5x| = 16 - 5x\\\\ \Rightarrow |2x - 4| + |3x - 6| + |16 - 5x| = \\\\ \Rightarrow (4 - 2x) + (6 - 3x) + (16 - 5x) = 26 - 10x\end{array}\)

برای هر \(x,y \in \mathbb{R}\)  داریم:

\(\left\{ \begin{array}{l}|x| - |y| \le |x + y| \le |x| + |y|\\\\|x| - |y| \le |x - y| \le |x| + |y|\end{array} \right.\)

\(\left\{ \begin{array}{l}x > 0,y > 0 \Rightarrow \frac{{|x|}}{x} = \frac{{|y|}}{y} = 1\\\\ \Rightarrow \frac{{|x|}}{x} + \frac{{|y|}}{y} = 1 + 1 = 2\\\\x < 0,y < 0 \Rightarrow \frac{{|x|}}{x} = \frac{{|y|}}{y} = - 1\\\\ \Rightarrow \frac{{|x|}}{x} + \frac{{|y|}}{y} = ( - 1) + ( - 1) = - 2\\\\x > 0,y < 0 \Rightarrow \frac{{|x|}}{x} + \frac{{|y|}}{y} = 1 + ( - 1) = 0\\\\x < 0,y > 0 \Rightarrow \frac{{|x|}}{x} + \frac{{|y|}}{y} = ( - 1) + 1 = 0\end{array} \right.\)

\(\begin{array}{l}\sqrt {{a^2} + 9 + \sqrt {36{a^2}} } = \sqrt {{a^2} + 9 + |6a|} \Rightarrow a > 0\\\\ \Rightarrow \sqrt {{a^2} + 9 + 6a} = |a + 3| = a + 3\end{array}\)