جواب تمرین صفحه 63 درس 4 ریاضی نهم (توان و ریشه)

 \({3^{ - 2}} = \frac{1}{{{3^2}}} = \frac{1}{9}\) الف

ب \({3^{ - 1}} = \frac{1}{{{3^1}}} = \frac{1}{3}\)

ج \({3^{ - 1}} \times {4^{ - 1}} = \frac{1}{{{3^1}}} \times \frac{1}{{{4^1}}} = \frac{1}{{{{12}^1}}} = {12^{ - 1}}\)

د \({3^{ - 1}} + {4^{ - 1}} = \frac{1}{{{3^1}}} + \frac{1}{{{4^1}}} = \frac{1}{3} + \frac{1}{4}\)

هـ \({5^{ - 2}} = \frac{1}{{{5^2}}} = \frac{1}{{25}}\)

و \({\left( { - 2} \right)^3} = \left( { - 2} \right) \times \left( { - 2} \right) \times \left( { - 2} \right) = - 8\)

\(\begin{array}{l}100\;kg = 100 \times 1000{\rm{\;}}g = {10^2} \times {10^3} = {10^5}\;g\\\\{10^5} \div {10^{ - 24}} = {10^{5 - \left( { - 24} \right)}} = {10^{29}}\end{array}\)

یعنی جرم یک وزنه 100 کیلوگرمی  \({10^{29}}\)  برابر جرم یک اتم هیدروژن است.

\({8^4} = {\left( {{2^3}} \right)^4} = {2^{12}}\;\;,\;\;\;{16^3} = {\left( {{2^4}} \right)^3} = {2^{12}} \Rightarrow {2^{11}} < {8^4} = {16^3}\)

  \({3^{ - 1}} > {3^{ - 2}}\) الف

ب \({2^0} > {2^{ - 5}}\)

ج \({(0/5)^{ - 2}} > {(0/6)^{ - 2}}\)

د \({5^{ - 1}} > 0\)

هـ \({\left( {\frac{{ - 8}}{{15}}} \right)^0} = 1\)

و \( - {5^{ - 2}} < {( - 5)^{ - 2}}\)

  الف

\(\begin{array}{l}{5^x} \times {5^{ - 3}} = {5^4}\\\\ \Rightarrow {5^{x + ( - 3)}} = {5^4}\\\\ \Rightarrow x - 3 = 4\\\\ \Rightarrow x = 7\end{array}\)

ب

\(\begin{array}{l}{5^x} \div {5^{ - 3}} = {5^4}\\\\ \Rightarrow {5^{x - ( - 3)}} = {5^4}\\\\ \Rightarrow x + 3 = 4\\\\ \Rightarrow x = 1\end{array}\)

گزینه های صحیح عبارتند از:

ب – ج – هـ - ح

به طور کلی نمی توانیم ادعا کنیم \({\left( {{a^m}} \right)^n} = {\left( {{a^n}} \right)^m}\)  است؛ زیرا:

\({\left( {{{\left( { - 3} \right)}^{\frac{1}{2}}}} \right)^2} \ne {\left( {{{\left( { - 3} \right)}^2}} \right)^{\frac{1}{2}}}\)

طرف اول (سمت چپ) در مجموعه ی اعداد حقیقی تعریف نشده است

\({\left( { - 3} \right)^{\frac{1}{2}}} = \sqrt { - 3} \notin R\)

زیرا عدد 3- در مجموعه اعداد حقیقی ریشه ی دوم ندارد

و سمت راست برابر 3 می شود

\({\left( {{{\left( { - 3} \right)}^2}} \right)^{\frac{1}{2}}} = {\left( { + 9} \right)^{\frac{1}{2}}} = \sqrt 9 = 3\)

الف

\(\begin{array}{l}{\left( {\frac{1}{3}} \right)^{ - 10}} \times {27^{ - 3}} = {3^{10}} \times {\left( {{3^3}} \right)^{ - 3}} = \\\\{3^{10}} \times {3^{ - 9}} = {3^1} = 3\end{array}\)

ب

\(\begin{array}{l}{\left( {0,2} \right)^{ - 4}} \times {25^{ - 2}} = {\left( {\frac{2}{{10}}} \right)^{ - 4}} \times {\left( {\frac{1}{{25}}} \right)^2} = \\\\{\left( {\frac{{10}}{2}} \right)^4} \times {\left( {\frac{1}{5}} \right)^4} = {\left( {5 \times \frac{1}{5}} \right)^4} = \\\\{1^4} = 1\end{array}\)

ج

\(\begin{array}{l}{\left( {\frac{{15}}{{14}}} \right)^{ - 4}} \times {\left( {\frac{{45}}{{28}}} \right)^4} = \\\\{\left( {\frac{{14}}{{15}}} \right)^4} \times {\left( {\frac{{45}}{{28}}} \right)^4} = \\\\{\left( {\frac{{14}}{{15}} \times \frac{{45}}{{28}}} \right)^4} = {\left( {\frac{3}{2}} \right)^4}\end{array}\)

د

\(\begin{array}{l}{\left( { - {5^{ - 2}}} \right)^{ - 1}} = {\left( { - \frac{1}{{{5^2}}}} \right)^{ - 1}} = \\\\{\left( { - \frac{1}{{25}}} \right)^{ - 1}} = {\left( { - 25} \right)^1} = - 25\end{array}\)

\(\begin{array}{l}{2^{ - 3}} = \frac{1}{8}\\\\{5^{ - 3}} = \frac{1}{{625}}\\\\{2^3} = 8\\\\{\left( { - \frac{1}{2}} \right)^{ - 2}} = 4\\\\{\left( { - 7} \right)^2} = 49\\\\{\left( { - 1} \right)^{21}} = - 1\\\\{1^{ - 90}} = 1\\\\ - {2^{ - 4}} = - \frac{1}{{16}}\\\\{\left( { - 1} \right)^{21}}\,,\, - {2^{ - 4}}\,,\,{5^{ - 3}}\,,\,{2^{ - 3}}\,,\,{1^{ - 90}}\,,\,{\left( { - \frac{1}{2}} \right)^{ - 2}}\,,\,{2^3}\,,\,{\left( { - 7} \right)^2}\end{array}\)

\(\begin{array}{l}{\left( {0,987} \right)^{10}} < {10^0} \Rightarrow {\left( {0,987} \right)^{10}} < 1\\\\1,2 > 1,02 \Rightarrow {\left( {1,2} \right)^7} > {\left( {1,02} \right)^7}\\\\\frac{5}{4} > 0,7 \Rightarrow {\left( {\frac{5}{4}} \right)^2} > {\left( {0,7} \right)^2}\\\\\frac{3}{4} = 0,75 \Rightarrow {\left( {\frac{3}{4}} \right)^2} > {\left( {0,75} \right)^3}\end{array}\)

عبارات صحیح از قرار زیر هستند:

\(\begin{array}{l}{\left( {0/987} \right)^{10}} < 10\\\\{\left( {\frac{3}{4}} \right)^2} = {\left( {0/75} \right)^3}\end{array}\)

الف

\(\begin{array}{l}\frac{{{{\left( {\frac{2}{3}} \right)}^3} \times {{\left( {\frac{8}{3}} \right)}^{ - 3}}}}{{ - {2^5} \times {2^{ - 8}}}} = \frac{{{{\left( {\frac{2}{3}} \right)}^3} \times {{\left( {\frac{3}{8}} \right)}^3}}}{{ - {2^{ - 3}}}} = \\\\\frac{{{{\left( {\frac{2}{3} \times \frac{3}{8}} \right)}^3}}}{{ - {2^{ - 3}}}} = \frac{{{{\left( {\frac{1}{3}} \right)}^3}}}{{ - {2^{ - 3}}}} = \\\\ - {2^3} \times {\left( {\frac{1}{2}} \right)^3} = {\left( { - \frac{1}{2}} \right)^3}\end{array}\)

ب

\(\begin{array}{l}{\left[ { - {{\left( {\frac{2}{3}} \right)}^{ - 2}}} \right]^{ - 1}} = {\left[ { - {{\left( {\frac{3}{2}} \right)}^2}} \right]^{ - 1}} = \\\\{\left( { - \frac{9}{4}} \right)^{ - 1}} = - \frac{4}{9}\end{array}\)