نمونه سوالات فصل 4 ریاضی نهم همراه با جواب تشریحی

\(\begin{array}{l}1){3^3} \times {4^0} - {5^3} \times 6 = 27 \times 1 - 125 \times 6 = \\\\27 - 750 = - 723\\\\2) - {5^2} + \frac{1}{{24}} + 300 \times {(\frac{1}{2})^2} = - 25 + \frac{1}{{16}} + 300 \times \frac{1}{4} = \\\\ - 25 + \frac{1}{{16}} + 75 = \frac{{ - 25 \times 16 + 1 + 75 \times 16}}{{16}} = \\\\\frac{{1 + 50 \times 16}}{{16}} = \frac{{801}}{{16}}\end{array}\)

\(\begin{array}{l}1){2^{ - 5}} = \frac{1}{{25}} = \frac{1}{{32}}\\\\2){(\frac{7}{3})^{ - 3}} = \frac{{{7^{ - 3}}}}{{{3^{ - 3}}}} = \frac{{\frac{1}{{{7^3}}}}}{{\frac{1}{{{3^3}}}}} = \frac{{{3^3}}}{{{7^3}}} = \frac{{27}}{{343}}\\\\3) - \frac{{{{( - 5)}^{ - 3}}}}{{ - {2^{ - 5}}}} = - \frac{{\frac{1}{{{{( - 5)}^3}}}}}{{ - \frac{1}{{{2^5}}}}} = - \frac{{\frac{1}{{ - 125}}}}{{ - \frac{1}{{32}}}} = - \frac{{32}}{{125}}\end{array}\)

\(\begin{array}{l}1){(\frac{2}{9})^5} \times {36^5} = \frac{{{2^5}}}{{{9^5}}} \times {36^5} = \frac{{{2^5} \times {{36}^5}}}{{{9^5}}} = \\\\{2^5} \times {(\frac{{36}}{9})^5} = {2^5} \times {4^5} = {2^5} \times {({2^2})^5} = {2^5} \times {2^{10}} = {2^{15}}\\\\2){(1\frac{1}{7})^3} \times {(\frac{{14}}{{16}})^{ - 2}} \times {(\frac{8}{7})^{ - 6}} = {(\frac{8}{7})^3} \times {(\frac{7}{8})^{ - 2}} \times {(\frac{8}{7})^{ - 6}} = \\\\{(\frac{8}{7})^3} \times {(\frac{8}{7})^2} \times {(\frac{8}{7})^{ - 6}} = {(\frac{8}{7})^{3 + 2 - 6}} = {(\frac{8}{7})^{ - 1}} = {(\frac{8}{7})^1}\\\\3){0/5^{ - 4}} \times {4^3} = {(\frac{1}{2})^{ - 4}} \times {({2^2})^3} = \\\\{2^{ + 4}} \times {2^6} = {2^{4 + 6}} = {2^{10}}\end{array}\)

\(\begin{array}{l}1)\frac{{{3^5} \times {9^{10}}}}{{{9^8} \times {3^4}}} = \frac{{{3^5} \times {3^{20}}}}{{{3^{16}} \times {3^4}}} = \frac{{{3^{5 + 20}}}}{{{3^{16 + 4}}}} = \frac{{{3^{25}}}}{{{3^{20}}}} = {3^{25 - 20}} = {3^5}\\\\2)\frac{{{{(\frac{7}{3})}^3}}}{{{{(2\frac{1}{3})}^4} \div {{(\frac{3}{7})}^{ - 2}}}} = \frac{{{{(\frac{7}{3})}^3}}}{{{{(\frac{7}{3})}^4} \div {{(\frac{7}{3})}^2}}} = {(\frac{7}{3})^{3 - (4 - 2)}} = \\\\{(\frac{7}{3})^{3 - 2}} = {(\frac{7}{3})^1}\end{array}\)

\(\frac{{9 \times {{10}^7}}}{{15}} = \frac{{90 \times {{10}^6}}}{{15}} = 6 \times {10^6}\)

\(\begin{array}{l}1)1/38 = 138 \times {10^{ - 2}} = 0/0138 \times {10^2}\\\\2)42 = 4200000 \times {10^{ - 5}} = 0/042 \times {10^3}\\\\3)40/01 = 4/001 \times {10^1} = 0/4001 \times {10^2}\end{array}\)

\(\begin{array}{l}1)4500000 = 4/5 \times {10^6}\\\\2)0/0080008100 = 8/00081 \times {10^{ - 3}}\\\\3)0/00000001001001 \times 9000 = 0/00009009009 = \\\\9/009009 \times {10^{ - 5}}\end{array}\)

\(\begin{array}{l}1)3/\overline 1 \times {10^{ - 4}} = 0/0003\overline 1 = 0/0003111 \cdots \\\\2)4/01 \times {10^3} = 4010\\\\3)5/00651 \times {10^{ - 4}} = 0/000500651\end{array}\)

\(\begin{array}{l}1)22 \times {10^{ - 4}} + 3/2 \times {10^{ - 3}} = 2/2 \times {10^{ - 3}} + 3/2 \times {10^{ - 3}} = \\\\ \Rightarrow 5/4 \times {10^{ - 3}}\\\\2)8/35 \times {10^{ - 4}} - 134 \times {10^{ - 4}} = - 125/65 \times {10^{ - 4}} = \\\\ \Rightarrow - 1/2565 \times {10^{ - 6}}\end{array}\)

الف)

\(\begin{array}{l}{R_A} = 265000km = 2/65 \times 2/65 \times {10^5}km\\\\{R_B} = 343000km = 3/43 \times {10^5}km\end{array}\)

ب)

\(\begin{array}{l}{R_B} - {R_A} = 3/43 \times {10^5} - 2/65 \times {10^5} = \\\\0/78 \times {10^5} = 7/8 \times {10^4}km\end{array}\)

\(\begin{array}{l}1)3\frac{4}{3} = \frac{{13}}{3} \Rightarrow {(3\frac{4}{3})^2} = \frac{{169}}{9}\\\\2) - {5^2} = - 25 \Rightarrow {( - {5^2})^2} = 625\\\\3){(\sqrt 8 )^2} = 9\\\\4)\frac{{32}}{{{{( - 2\sqrt 2 )}^2}}} = \frac{{32}}{8} = 4 \Rightarrow {(\frac{{32}}{{{{( - 2\sqrt 2 )}^2}}})^2} = 16\end{array}\)

\(\begin{array}{l}1) - \sqrt {1/21} = - 1/1\\\\2)3\sqrt {0/64} - 2\sqrt {1/44} = \\\\ \Rightarrow 3 \times 0/8 - 2 \times 1/2 = 2/4 - 2/4 = 0\\\\3)2\sqrt {\frac{{81}}{{121}}} - 3\sqrt {\frac{{25}}{{64}}} = 2 \times \frac{9}{{11}} - 3 \times \frac{5}{8} = \frac{{18}}{{11}} - \frac{{15}}{8} = \\\\ \Rightarrow \frac{{144 - 165}}{{88}} = - \frac{{21}}{{88}}\end{array}\)

\(\begin{array}{l}1)3\sqrt[3]{{ - 1331}} = 3( - 11) = - 33\\\\2){(\frac{5}{2})^3} = \frac{{125}}{8}\\\\3)2\sqrt[3]{{512}} - 5\sqrt[3]{{1331}} = 2(8) - 5(11) = 16 - 55 = - 39\\\\4)\sqrt[3]{{\frac{{125}}{{0/001}}}} = \frac{5}{{0/1}} = 50\end{array}\)

\(\begin{array}{l}1)\sqrt {{x^2}} + \sqrt {{y^2}} = |x| + |y| = - x + y\\\\2)x < 2 \Rightarrow 2 - x > 0 \Rightarrow \\\\ \Rightarrow \sqrt[2]{{{{(2 - x)}^2}}} - \sqrt[3]{{{{(x - 2)}^3}}} = \\\\ \Rightarrow |2 - x| - (x - 2) = 2 - x - x + 2 = 4 - 2x\\\\3)y > 0,x > y \Rightarrow x - y > 0\\\\ \Rightarrow \sqrt {({x^2} + {y^2} - 2xy)} - \sqrt {{x^4}} = \sqrt {{{(x - y)}^2}} - {x^2} = \\\\ \Rightarrow |x - y| - {x^2} = x - y - {x^2}\end{array}\)

\({a^3} = 0/001331 = {(0/11)^3} \Rightarrow a = 0/11m\)

الف) قطر یک وجه

\(\begin{array}{l}\sqrt {{a^2} + {a^2}} = \sqrt {2{a^2}} = \sqrt 2 \times \sqrt {{a^2}} = \sqrt 2 a = \\\\\sqrt 2 \times 0/11 = 0/11\sqrt 2 m\end{array}\)

ب) مساحت کل مکعب

\(6 \times {a^2} = 6{a^2} = 6 \times {(0/11)^2} = 6 \times 0/0121 = 0/0726{m^2}\)

\(\begin{array}{l}1)\sqrt {165 - 3\sqrt {49} } = \sqrt {165 - 3 \times 7} = \sqrt {165 - 21} = \\\\ \Rightarrow \sqrt {144} = 12\\\\2)\sqrt[3]{{125}} - \sqrt[3]{{1/1331}} = 5 - 1/1 = 3/9\\\\3)\sqrt {\sqrt {256} + \sqrt {81} } = \sqrt {16 + 9} = \sqrt {25} = 5\end{array}\)

\(\begin{array}{l}3\sqrt {98} + 5\sqrt {32} - 4\sqrt {338} = \\\\ \Rightarrow 3\sqrt {49 \times 2} + 5\sqrt {16 \times 2} - 4\sqrt {169 \times 2} = \\\\ \Rightarrow 3 \times 7\sqrt 2 + 5 \times 4\sqrt 2 - 4 \times 13\sqrt 2 = \\\\21\sqrt 2 + 20\sqrt 2 - 52\sqrt 2 = (21 + 20 - 52)\sqrt 2 = - 11\sqrt 2 \end{array}\)

\(\begin{array}{l}1){(\sqrt 2 - \sqrt 3 )^2} = {(\sqrt 2 )^2} + {(\sqrt 3 )^2} - 2(\sqrt 2 )(\sqrt 3 ) = \\\\ \Rightarrow 2 + 3 - 2\sqrt 6 = 5 - 2\sqrt 6 \\\\2)(\sqrt[3]{5} - \sqrt[3]{2})(\sqrt[3]{5} + \sqrt[3]{2}) = {(\sqrt[3]{5})^2} - {(\sqrt[3]{3})^2} = \\\\ \Rightarrow \sqrt[3]{{{5^2}}} - \sqrt[3]{{{2^2}}} = \sqrt[3]{{25}} - \sqrt[3]{6}\end{array}\)

\({x^2} = {(\sqrt 2 )^2} + {(\sqrt 5 )^2} = 2 + 5 = 7 \Rightarrow x = \sqrt 7 \)

\(\begin{array}{l}1)\frac{3}{{\sqrt 5 }} + \frac{{4\sqrt 2 }}{{3\sqrt 3 }} = \frac{3}{{\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }} + \frac{{4\sqrt 2 }}{{3\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} = \\\\ \Rightarrow \frac{{3\sqrt 5 }}{5} + \frac{{4\sqrt 6 }}{9} = \frac{{27\sqrt 5 + 20\sqrt 6 }}{{45}}\\\\2)\frac{{5\sqrt 2 }}{{\frac{5}{{\sqrt 2 }}}} - \frac{{5\sqrt {18} }}{{3\sqrt 6 }} = \frac{{5\sqrt 2 }}{{\frac{5}{{\sqrt 2 }}}} \times \frac{{\frac{1}{{\sqrt 2 }}}}{{\frac{1}{{\sqrt 2 }}}} - \frac{{5\sqrt {18} }}{{3\sqrt 6 }} \times \frac{{\sqrt 6 }}{{\sqrt 6 }} = \\\\ \Rightarrow \frac{5}{{\frac{5}{2}}} - \frac{{5\sqrt {72} }}{{3 \times 6}} = 2 - \frac{{5 \times 6\sqrt 2 }}{{18}} = 2 - \frac{{5\sqrt 2 }}{3} = \frac{{6 - 5\sqrt 2 }}{3}\end{array}\)

\(\frac{{ - 1}}{{27}}\)

1

\(\frac{1}{{256}}\)

\({10^{26}}\)

\(\,[1,10)\,\,\,\,\,\,\,\,\,\,\, - a \times {10^n}\)

\( - a \times {10^n}\)

منفی

\(7/631005 \times {10^3}\)

\( - 1/08329 \times {10^{ - 2}}\)

\(\frac{1}{{24}}\)

\(y \le 0,x \ge 0\)

\({a^3}\)

وجود ندارد.

6

6-

3+x

27

\(5 - 2\sqrt 6 \)

برابر \(\frac{1}{4}\)  می باشد.

نصف \({16^{ - 6}}\)  برابر \({8^{ - 6}}\)  نمی باشد.

برابر \(\frac{1}{{xy}}\)  می باشد.

برابر \({9^{ - 4}}\)  می باشد.

برابر هستند.

هر عدد مثبت دارای دو ریشه دوم است. عدد صفر، فقط یک ریشه دوم دارد که برابر صفر می باشد.

فرجه جذر برابر 2 می باشد.

به ازای جمیع مقادیر نامنفی برقرار است؛ یعنی \(x \ge 0\)

برابر با \(30\sqrt 2 \)  می باشد.

\(\frac{{{{( - 2)}^{ - 18}}}}{2} = \frac{{{2^{ - 18}}}}{2} = {2^{ - 19}}\)

روش اول: اگر تعداد رقم های یک عدد روج باشد، تعداد رقم های ریشه دوم آن نصف می شود.

روش دوم: فرض کنید عدد 16 رقمی، x باشد؛ در این صورت داریم:

\(\begin{array}{l}1 \times {10^{15}} \le x < 1 \times {10^{16}} \Rightarrow 10 \times {10^{14}} \le x < 1 \times {10^{16}}\\\\ \Rightarrow 3/16 \times {10^7} \le \sqrt x < 1 \times {10^8} \Rightarrow 3 \times {10^7} < x < 1 \times {10^8}\end{array}\)

در این صورت چون طبق داده مسأله می دانیم عدد x جذر کامل دارد؛ یعنی \(\sqrt x \)  دارای عدد اعشاری نمی باشد؛ بنابراین \(\sqrt x \)  حتما 8 رقمی می باشد.

\(\begin{array}{l}{9^{5n - 10}} = 1 = {9^0} \Rightarrow 5n - 10 = 0 \Rightarrow 5n = 10\\\\ \Rightarrow n = \frac{{10}}{5} \Rightarrow n = 2\\\\2(2{n^{ - 2}} - 3{n^{ - 1}}) = 2(2{(2)^{ - 2}} - 3{(2)^{ - 1}}) = \\\\2(2 \times \frac{1}{4} - 3 \times \frac{1}{2}) = 2(\frac{1}{2} - \frac{3}{2}) = - 2\end{array}\)

\(\begin{array}{l}{( - 1)^1} \times 1 + {( - 1)^2} \times 2 + {( - 1)^3} \times 3 + \cdots {( - 1)^{200}} \times 200 = \\\\ - 1 + 2 - 3 + 4 + \cdots - 199 + 200 = \\\\ \Rightarrow ( - 1 + 2) + ( - 3 + 4) + \cdots + ( - 199 + 200) = \\\\ \Rightarrow 1 + 1 + \cdots + 1 = 1 \times 100 = 100\end{array}\)

\(\begin{array}{l}{1^2} - {2^2} + {3^2} - {4^2} + \cdots + {9^2} - {10^2} = \\\\ \Rightarrow ({1^2} + {2^2}) + ({3^2} - {4^2}) + \cdots + ({9^2} - {10^2}) = \\\\ \Rightarrow ((1 - 2)(1 + 2)) + ((3 - 4)(3 + 4)) + \cdots \\\\ \Rightarrow \cdots + ((9 - 10) + (9 + 10)) = \\\\ \Rightarrow ( - 1)(3) + ( - 1)(7) + \cdots + ( - 1)(19) = \\\\ \Rightarrow - 3 - 7 - 11 - 15 - 19 = - (3 + 7 + 11 + 15 + 19) = \\\\ \Rightarrow - 55\end{array}\)

\(873/24 \times {10^{ - 5}} = 8/7324 \times {10^{ - 3}}\)

\(\begin{array}{l}\frac{{0/003 \times 0/0144}}{{12 \times {{10}^{ - 5}}}} = \frac{{3 \times {{10}^{ - 3}} \times 1/44 \times {{10}^{ - 2}}}}{{1/2 \times {{10}^{ - 4}}}} = \\\\3 \times 1/2 \times {10^{ - 1}} = 3/6 \times {10^{ - 1}}\end{array}\)

\(\begin{array}{l}0/013 \times 0/0003 = 1/3 \times {10^{ - 2}} \times 3 \times {10^{ - 4}} = 3/9 \times {10^{ - 6}}\\\\ \Rightarrow 3/9 \times {10^{ - 6}} < 4 \times {10^n} \Rightarrow n = - 6, - 5, - 4, \cdots \end{array}\)

کوچک ترین عدد صحیح که می تواند به جای n قرار بگیرد، عدد 6- می باشد.

\({810^{ - 9}} - 1/8 \times {10^{ - 6}} = 0/81 \times {10^{ - 6}} - 1/8 \times {10^{ - 6}} = - 1 \times {10^{ - 6}}\)

\(\begin{array}{l}0/00000103 = 1/03 \times {10^{ - 6}} \Rightarrow \left\{ \begin{array}{l}a = 1/03\\\\n = - 6\end{array} \right. \Rightarrow a - n = \\\\ \Rightarrow 1/03 + 6 = 7/03\end{array}\)

\(\sqrt {{{64}^3} \times {{216}^2} \times {5^6}} = \sqrt {{8^6} \times {6^6} \times {5^6}} = \sqrt {{{240}^6}} = {240^3}\)

اگر \(x \in (0,1)\)  باشد، جذر آن از خودش بزرگ تر و توان 2 آن از خودش کوچک تر است؛ به عبارتی:

\(x \in (0,1) \Rightarrow {x^2} < x < \sqrt x \)

\(\sqrt {{2^{{4^{{{( - 2)}^{( - 2)}}}}}}} = \sqrt {{2^{{4^{(\frac{1}{4})}}}}} = \sqrt 2 \)

\(\begin{array}{l}\sqrt {10 + 2\sqrt {21} } = \sqrt {7 + 3 + 2\sqrt {7 \times 3} } = \\\\\sqrt {{{(\sqrt 7 + \sqrt 3 )}^2}} = \sqrt 7 + \sqrt 3 \end{array}\)

\(\begin{array}{l}\sqrt {\frac{5}{{\sqrt {30 - 5} }}} = \sqrt {\frac{5}{{\sqrt {30} - 5}} \times \frac{{\sqrt {30 + 5} }}{{\sqrt {30} + 5}}} = \\\\\sqrt {\frac{{5(\sqrt {30} + 5)}}{{30 - 25}}} = \sqrt {\frac{{5(\sqrt {30} + 5)}}{5}} = \sqrt {\sqrt {30} + 5} \end{array}\)

\(\sqrt[3]{{\frac{{625}}{5}}} - \sqrt[3]{{216}} = \sqrt[3]{{125}} - \sqrt[3]{{216}} = 5 - 6 = - 1\)

\(\frac{6}{{\sqrt[3]{{36}}}} \times \frac{{\sqrt[3]{6}}}{{\sqrt[3]{6}}} = \frac{{6\sqrt[3]{6}}}{{\sqrt[3]{{36 \times 6}}}} = \frac{{6\sqrt[3]{6}}}{{\sqrt[3]{{216}}}} = \frac{{6\sqrt[3]{6}}}{6} = \frac{{\sqrt[3]{6}}}{1}\)

\(\begin{array}{l}\frac{1}{{\sqrt x }} - \frac{1}{{\sqrt x - \sqrt y }} = \\\\ \Rightarrow \frac{1}{{\sqrt x }} \times \frac{{\sqrt x }}{{\sqrt x }} - \frac{1}{{\sqrt x - \sqrt y }} \times \frac{{\sqrt x + \sqrt y }}{{\sqrt x + \sqrt y }} = \\\\ \Rightarrow \frac{{\sqrt x }}{x} - \frac{{\sqrt x + \sqrt y }}{{x + y}} = \frac{{\sqrt x (x + y) - (\sqrt x + \sqrt y )x}}{{x(x + y)}} = \\\\ \Rightarrow \frac{{x\sqrt x + y\sqrt x - x\sqrt x }}{{x(x + y)}} = \frac{{y\sqrt x }}{{x(x + y)}}\end{array}\)

\(\begin{array}{l}\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \sqrt {48} = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} - \sqrt {48} = \\\\ \Rightarrow \frac{{{{(\sqrt 3 + \sqrt 2 )}^2}}}{{3 - 2}} - \sqrt {48} = \frac{{3 + 2 + 12\sqrt 2 - \sqrt {48} }}{1} = \\\\ \Rightarrow 5 + 12\sqrt 2 - 4\sqrt 3 \end{array}\)

\(\begin{array}{l}\frac{{8\sqrt 3 }}{{\sqrt 3 + \sqrt {12} + \sqrt {75} }} = \frac{{8\sqrt 3 }}{{\sqrt 3 + 2\sqrt 3 + 5\sqrt 3 }} = \\\\ \Rightarrow \frac{{8\sqrt 3 }}{{8\sqrt 3 }} = 1\end{array}\)

\( - 0/2\)