نمونه سوالات تشریحی فصل 7 ریاضی هشتم همراه با جواب تشریحی

\(\begin{array}{l}1)\frac{{{{( - 11)}^0} - {{( - 3)}^3} - {{( - 2)}^5}}}{{{{( - 1)}^7} - {{( - 1)}^{11}} - {{(1)}^0}}} = \frac{{1 - ( - 27) - ( - 32)}}{{ - 1 - ( - 1) - 1}} = \\\\\frac{{1 + 27 + 32}}{{ - 1 + 1 - 1}} = \frac{{60}}{{ - 1}} = - 60\\\\2)3 \times {2^3} - 2 \times {( - 3)^2} = 3 \times 8 - 2 \times 9 = 24 - 18 = 6\\\\3){(x)^2} - ( - {x^2}) = {x^2} + {x^2} = 2{x^2}\end{array}\)

\(\begin{array}{l}1)6 \times {3^2} + 3 \times {3^2} = {3^2} \times (6 + 3) = {3^2} \times 9 = 9 \times 9 = 81 = {3^4}\\\\2){6^5} + {6^5} + {6^5} + {6^5} + {6^5} + {6^5} = {6^5}(1 + 1 + 1 + 1 + 1 + 1) = \\\\{6^5} \times 6 = {6^{5 + 1}} = {6^6}\\\\3)9 \times {4^3} + 5 \times {4^3} - 2 \times {4^3} - 5 \times {4^3} = \\\\{4^3} \times (9 + 5 - 2 - 5) = 7 \times {4^3}\end{array}\)

\(\begin{array}{l}1){3^5} \times {3^6} \times {2^7} \times {2^4} = {3^{5 + 6}} \times {2^7} \times {({2^2})^2} = \\\\{3^{11}} \times {2^7} \times {2^4} = {3^{11}} \times {2^{11}} = {(3 \times 2)^{11}} = {6^{11}}\\\\2)4 \times {9^2} - {9^2} = {9^2} \times (4 - 1) = {9^2} \times 3 = ({3^2}) \times 3 = \\\\{3^4} \times 3 = {3^{4 + 1}} = {3^5}\\\\3){x^5} \times {x^4} \div {( - {x^3})^2} = {x^{5 + 4}} \div {x^6} = {x^{9 - 6}} = {x^3}\\\\4){( - {x^2})^3} \times {( - x)^5} \div {x^4} = ( - {x^6}) \times ( - {x^5}) \div {x^4} = \\\\{x^{6 + 5 - 4}} = {x^7}\end{array}\)

\(\begin{array}{*{20}{l}}{1)\frac{{{{( - a)}^3} \times {{( - {a^3})}^2}}}{{{{({a^5})}^2} \times {{( - {a^2})}^3}}} = \frac{{( - {a^3}) \times ({a^6})}}{{{a^{10}} \times ( - {a^6})}} = \frac{{ - {a^{3 + 6}}}}{{ - {a^{10 + 6}}}} = \frac{{{a^9}}}{{{a^{16}}}} = }\\{\frac{1}{{{a^7}}} = {a^{ - 7}}}\\{}\\{2)\frac{{0/12 \times {{10}^3} + 0/24 \times {{10}^6}}}{{0/06 \times {{10}^5}}} = \frac{{(0/12 \times {{10}^3}) \times (1 + 2 \times {{10}^3})}}{{0/06 \times {{10}^5}}} = }\\{}\\{\frac{{2 \times (2001)}}{{{{10}^2}}} = \frac{{4002}}{{{{10}^2}}} = 40/02 = 4/002 \times {{10}^{ + 1}}}\end{array}\)

\(\begin{array}{l}1){6^{12}} \div {({6^3})^2} = {6^{12}} \div {6^6} = {6^{12 - 6}} = {6^6}\\\\2)4 \times {2^2} \times \frac{1}{{16}} = {2^2} \times {2^2} \times \frac{1}{{{2^4}}} = {2^2} \times {2^2} \times {2^{ - 4}} = {2^{2 + 2 - 4}} = {2^0} = 1\\\\3)\frac{{3 \times {2^8} + 5 \times {2^8}}}{{3 \times {4^2} + 16}} = \frac{{{2^8}(3 + 5)}}{{3 \times {4^2} + {4^2}}} = \frac{{{2^8} \times 8}}{{{4^2}(3 + 1)}} = \\\\\frac{{{2^8} \times {2^3}}}{{{{({2^2})}^2} \times 4}} = \frac{{{2^{8 + 3}}}}{{{2^4} \times {2^2}}} = \frac{{{2^{11}}}}{{{2^{4 + 2}}}} = \frac{{{2^{11}}}}{{{2^6}}} = {2^{11 - 6}} = {2^5}\end{array}\)

\(\begin{array}{l}1)\frac{{125 + 125 + 125 + 125 + 125}}{{8 + 8}} = \frac{{125 \times 5}}{{8 \times 2}} = \frac{{{5^3} \times 5}}{{{2^3} \times 2}} = {(\frac{5}{2})^4}\\\\2)\frac{{{9^x} + {9^x} + {9^x}}}{{{3^{x - 2}} \times 2 + {3^{x - 2}}}} = \frac{{3 \times {9^x}}}{{{3^{x - 2}}(2 + 1)}} = \frac{{3 \times {{({3^2})}^x}}}{{3 \times {3^{x - 2}}}} = \frac{{{3^{2x}}}}{{{3^{x - 2}}}} = \\\\{3^{2x - (x - 2)}} = {3^{x + 2}}\\\\3)\frac{{{3^6} + {3^7} + {3^8}}}{{9 + 3 + 1}} = \frac{{{3^6}(1 + 3 + {3^2})}}{{({3^{}}8}}\end{array}\)

\(\begin{array}{l}1){2^x} = 5 \Rightarrow {8^x} = {({2^3})^x} = {2^{3x}} = {({2^x})^3} = {5^3} = 125\\\\2){6^x} = 7 \Rightarrow {216^x} + {6^{x + 2}} = {({6^3})^x} + {6^x} \times {6^2} = \\\\{6^{3x}} + 7 \times 36 = {({6^x})^3} + 7 \times 36 = {7^3} + 7 \times 36 = \\\\7({7^2} + 36) = 7(49 + 36) = 7 \times 85 = 595\end{array}\)

\(\begin{array}{l}1){3^9},{9^2},{81^2} \Rightarrow {3^5},{9^2} = {({3^2})^2} = {3^4},{81^2} = {({3^4})^2} = {3^8}\\\\ \Rightarrow {9^2} < {3^5} < {81^2} \Rightarrow {9^2},{3^5},{81^2}\\\\2){128^{200}},{64^{234}},{512^{150}} \Rightarrow {128^{200}} = {({2^7})^{200}} = {2^{1400}}\\\\{64^{234}} = {({2^6})^{234}} = {2^{2004}},{512^{150}} = {({2^9})^{150}} = {2^{1350}}\\\\ \Rightarrow {512^{150}} < {128^{200}} < {64^{234}} \Rightarrow {512^{150}},{128^{200}},{64^{234}}\end{array}\)

\(\begin{array}{l}\frac{{16A \times 81B}}{{8A \times 27B}} = \frac{{{2^4} \times {2^4} \times {3^4} \times {3^4}}}{{{2^3} \times {2^4} \times {3^3} \times {3^4}}} = \frac{{{2^8} \times {3^8}}}{{{2^7} \times {3^7}}} = \frac{{{{(2 \times 3)}^8}}}{{{{(2 \times 3)}^7}}} = \\\\ = \frac{{{6^8}}}{{{6^7}}} = {6^{8 - 7}} = {6^1} = 6\end{array}\)

\(\begin{array}{l}1)\sqrt {196} - \sqrt {225} + \sqrt 9 = 14 - 15 + 3 = 2\\\\2)\sqrt {\sqrt {100} + \sqrt {49} - } \sqrt {64} = \sqrt {10 + 7 - 8} = \sqrt 9 = 3\end{array}\)

\(\begin{array}{l}1)\sqrt {1/21} - \sqrt {0/81} = 1/1 - 0/9 = 0/2\\\\2)\sqrt {1 - \frac{8}{9}} + \sqrt {1 + \frac{{44}}{{81}}} = \sqrt {\frac{{9 - 8}}{9}} + \sqrt {\frac{{81 + 44}}{{81}}} = \\\\\sqrt {\frac{1}{9}} + \sqrt {\frac{{125}}{{81}} = } \frac{1}{3} + \frac{5}{9} = \frac{3}{9} + \frac{5}{9} = \frac{8}{9}\end{array}\)

\(\begin{array}{l}1)\sqrt 2 + \sqrt 8 + \sqrt {50} = \sqrt 2 + \sqrt {4 \times 2} + \sqrt {25 \times 2} = \\\\\sqrt 2 + 2\sqrt 2 + 5\sqrt 2 = \sqrt 2 (1 + 2 + 5) = 8\sqrt 2 \\\\2)\sqrt {20} - \sqrt 5 + \sqrt {605} = \sqrt {4 \times 5} - \sqrt 5 + \sqrt {121 \times 5} = \\\\2\sqrt 5 - \sqrt 5 + 11\sqrt 5 = \sqrt 5 (2 - 1 + 11) = 12\sqrt 5 \end{array}\)

\(\begin{array}{l}1)\sqrt {2\sqrt {14} \times 7\sqrt {14} } = \sqrt {14 \times {{(\sqrt {14} )}^2}} = \sqrt {14 \times 14} = \sqrt {{{14}^2}} = 14\\\\2)(\sqrt {27} + \sqrt {12} )\sqrt 3 = \sqrt {27} \times \sqrt 3 + \sqrt {12} \times \sqrt 3 = \\\\\sqrt {27 \times 3} + \sqrt {12 \times 3} = \sqrt {81} + \sqrt {36} = 9 + 6 = 15\\\\3)\sqrt {{3^5}} + \sqrt {27} = \sqrt {{3^4} \times 3} \sqrt {{3^2} \times 3} = {3^2}\sqrt 3 + 3\sqrt 3 = \\\\9\sqrt 3 + 3\sqrt 3 = 12\sqrt 3 \\\\4)\sqrt {108} - \sqrt {75} + \sqrt 3 = \sqrt {36 \times 3} - \sqrt {25 \times 3} + \sqrt 3 = \\\\6\sqrt 3 - 5\sqrt 3 + \sqrt 3 = \sqrt 3 (6 - 5 + 1) = 2\sqrt 3 \end{array}\)

\(\begin{array}{l}1)\frac{{4\sqrt {10} }}{{\sqrt {20} \times \sqrt 8 }} = \frac{{4\sqrt {10} }}{{\sqrt {160} }} = \frac{{4\sqrt {10} }}{{\sqrt {16 \times 10} }} = \frac{{4\sqrt {10} }}{{4\sqrt {10} }} = 1\\\\2)\frac{{\sqrt {3000} }}{{\sqrt {270} }} = \frac{{\sqrt {30 \times 100} }}{{\sqrt {30 \times 9} }} = \frac{{10\sqrt {30} }}{{3\sqrt {30} }} = \frac{{10}}{3}\end{array}\)

\(\begin{array}{l}1)2\sqrt x - 5 = 3 \Rightarrow 2\sqrt x = 3 + 5 = 8 \Rightarrow \sqrt x = \frac{8}{2} = 4\\\\ \Rightarrow x = {4^2} \Rightarrow x = 16\\\\2)\sqrt {{x^3}} = 2\sqrt 2 \Rightarrow \sqrt {{x^3}} = \sqrt 4 \sqrt 2 = \sqrt 8 = \sqrt {{2^3}} \Rightarrow x = 2\end{array}\)

\(\begin{array}{l}1)\sqrt {27} \Rightarrow \sqrt {25} < \sqrt {27} < \sqrt {36} \Rightarrow 5 < \sqrt {27} < 6\\\\2)\sqrt {179} \Rightarrow \sqrt {169} < \sqrt {179} < \sqrt {196} \Rightarrow 13 < \sqrt {179} < 14\\\\3)\sqrt {288} \Rightarrow \sqrt {256} < \sqrt {288} < \sqrt {289} \Rightarrow 16 < \sqrt {288} < 17\end{array}\)

\(1)\sqrt {54} \Rightarrow \sqrt {49} < \sqrt {54} < \sqrt {64} \Rightarrow 7 < \sqrt {54} < 8\)

\( \Rightarrow \sqrt {54} \simeq 7/3\)

\(2)\sqrt {120} \Rightarrow \sqrt {100} < \sqrt {120} < \sqrt {121} \Rightarrow 10 < \sqrt {120} < 11\)

\( \Rightarrow \sqrt {120} \simeq 10/9\)

روش به دست آوردن بدین صورت است که جدول را به صورتی پر می کنیم که در نهایت، عددی که در ردیف مجذور بدست می آید، بزرگترین عدد کوچکتر از عدد زیر رادیکال باشد.

\(\begin{array}{l}1)3 + \sqrt {26} \\\\26 = 25 + 1 = {5^2} + {1^2}\end{array}\)

\(\begin{array}{l}2)2 - \sqrt {18} \\\\18 = 16 + 1 + 1 = {4^2} + {1^2} + {1^2}\end{array}\)

\(1) \Rightarrow 2 - \sqrt 2 \)

\(2) \Rightarrow - 3 - \sqrt 3 \)

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