نمونه سوالات تستی فصل 7 ریاضی هشتم همراه با جواب تشریحی

\(\left. \begin{array}{l}{23^0} = 1\\\\{2^4} = 16\\\\{( - 3)^2} = 9\\\\{4^2} = 16\end{array} \right\} \Rightarrow {23^0} + {2^4} + {( - 3)^2} - {4^2} = 1 + 16 + 9 - 16 = 10\)

\(\left. \begin{array}{l}{1^{98}} = 1\\\\{( - 3)^2} = 9\\\\{5^1} = 5\end{array} \right\} \Rightarrow {1^{98}} - {( - 3)^2} + {5^1} = 1 - 9 + 5 = - 3\)

\(\begin{array}{l}\left. \begin{array}{l}{9^8} = {({3^2})^8} = {3^{16}}\\\\{27^2} = {({3^3})^2} = {3^6}\\\\{9^7} = {({3^2})^7} = {3^{14}}\end{array} \right\}\frac{{{3^8} + {9^8}}}{{{{27}^2} + {9^6}}} = \frac{{{3^8} + {3^{16}}}}{{{3^6} + {3^{12}}}} = \frac{{{3^8}(1 + {3^8})}}{{{3^6}(1 + {3^8})}} = \\\\ \Rightarrow {3^{8 - 6}} = {3^2} = 9\end{array}\)

\(\begin{array}{l}\frac{{{9^{3x - 5y + 4}}}}{{{{27}^{2x - 2y + 1}}}} = \frac{{{{({3^2})}^{3x - 5y = 4}}}}{{{{({3^2})}^{2x - 2y + 1}}}} = \frac{{{3^{6x}} - 10y + 8}}{{{3^{6x - 6y + 3}}}} = \\\\{3^{(6x - 10y + 8) - (6x - 6y + 3)}} = {3^{ - 4y + 5}} = 27 = {3^3} \Rightarrow - 4y + 5 = 3\\\\ \Rightarrow - 4y = - 2 \Rightarrow y = \frac{1}{2}\end{array}\)

هر یک از پرانتز ها را به توان می رسانیم، سپس با توجه به قوانین ضرب اعداد توان دار، حاصل را محاسبه می کنیم:

\(\begin{array}{l}{({x^3}{y^2})^2} = {x^6}{y^4}\\\\{( - 3{x^2}y)^3} = - 27{x^6}{y^3}\\\\ \Rightarrow 3{({x^3}{y^2})^2} \times {( - 3{x^2}y)^3} = 3{x^6}{y^4} \times ( - 27{x^6}{y^3}) = \\\\ - 81{x^{12}}{y^7}\end{array}\)

\(\begin{array}{l}{3^{x + 2}} + {3^{x + 1}} + {3^x} = 117 \Rightarrow {3^x}({3^2} + 3 + 1) = 117\\\\{3^x} \times (9 + 3 + 1) = 117 \Rightarrow {3^x} \times 13 = 117\\\\ \Rightarrow {3^x} = \frac{{117}}{{13}} = 9 = {3^2} \Rightarrow x = 2\end{array}\)

\(\begin{array}{l}\frac{{4 \times {3^{x + 1}} + {3^{x - 1}} - {3^{x\_3}}}}{{{3^{x - 1}}}} = \frac{{{3^{x - 1}}(4 \times {3^2} + 1 - {3^3})}}{{{3^{x - 1}}}} = \\\\4 \times 9 + 1 - 27 = 10\end{array}\)

دقت کنید که در این سوال از کوچکترین توان یعنی \({3^{x - 1}}\)  در صورت کسر فاکتور گرفته ایم.

\({2^a} \times {3^b} = 648 = 8 \times 81 = {2^3} \times {3^4} \Rightarrow \left\{ \begin{array}{l}a = 3\\\\b = 4\end{array} \right. \Rightarrow \frac{a}{b} = \frac{3}{4}\)

\({( - {( - 3)^3})^{ - 2}} = {( - ( - {3^2}))^{ - 2}} = {({3^3})^{ - 2}} = {3^{ - 6}}\)

قرینه معکوس \({3^{ - 6}}\)  برابر \( - {3^6}\)  می باشد که در نهایت برابر 729- می شود.

\({24^x} = {({2^3} \times 3)^x} = {2^{3x}} \times {3^x} = {({2^x})^3} \times {3^x} = {a^3} \times b = {a^3}b\)

\({({3^5})^2} = {3^{5 \times 2}} = {3^{10}}\)

\({( - 3)^2} \div {( - \frac{1}{2})^3} = 9 \div \frac{{ - 1}}{8} = 9 \times ( - 8) = - 72\)

\(\begin{array}{l}{3^8} \div {( - 27)^3} = - ({3^8} \div {27^3}) = - ({3^8} \div {({3^3})^3}) = \\\\ - ({3^8} \div {3^9}) = - {3^{8 - 9}} = - {3^{ - 1}} = - \frac{1}{3}\end{array}\)

\({( - {3^2})^3} = - {({3^2})^3} = - {3^6}\)

\(\sqrt {\frac{{25}}{{36}} \times 1 \times \frac{9}{{25}}} = \sqrt {\frac{{25}}{{36}} \times \frac{9}{{25}}} = \sqrt {\frac{9}{{36}}} = \sqrt {\frac{1}{4}} = \frac{1}{2}\)

\(\begin{array}{l}\frac{{\sqrt 2 + 2 + \sqrt 6 + \sqrt 8 }}{{\sqrt 5 + \sqrt {10} + \sqrt {15} + \sqrt {20} }} = \frac{{\sqrt 2 (1 + \sqrt 2 + \sqrt 3 + 2)}}{{\sqrt 5 (1 + \sqrt 2 + \sqrt 3 + 2)}} = \\\\\frac{{\sqrt 2 }}{{\sqrt 5 }} = \sqrt {\frac{2}{5}} \end{array}\)

\(\begin{array}{l}\frac{{\sqrt {0/0196} }}{{\sqrt {0/04} }} - \frac{{\sqrt {0/008} }}{{\sqrt {0/2} }} = \sqrt {\frac{{0/0196}}{{0/04}}} - \sqrt {\frac{{0/008}}{{0/2}}} = \\\\\sqrt {0/49} - \sqrt {0/04} = 0/7 - 0/2 = 0/5 = \frac{1}{2}\end{array}\)

\(\begin{array}{l}\sqrt {0/36} \div (\sqrt {0/01} + \sqrt {0/04} + \sqrt {0/09} ) = \\\\0/6 \div (0/1 + 0/2 + 0/3) = 0/6 \div 0/6 = 1\end{array}\)

\(\frac{3}{2} = \frac{{12}}{{\sqrt {2a} }} \Rightarrow \sqrt {2a} = \frac{{2 \times 12}}{3} = 8 \Rightarrow 2a = {8^2} = 64 \Rightarrow a = 32\)

\(\sqrt {{3^{37}} \times 27} = \sqrt {{3^{37}} \times {3^3}} = \sqrt {{3^{37 + 3}}} = \sqrt {{3^{40}}} = {3^{20}}\)