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Câu hỏi: Thực hiện phép tính:
a) \({81^{ - 0,75}} + {\left( {{1 \over {125}}} \right)^{{{ - 1} \over 3}}} - {\left({{1 \over {32}}} \right)^{{{ - 3} \over 5}}};\)
b) \(0,{001^{{{ - 1} \over 3}}} - {\left( { - 2} \right)^{ - 2}}{. 64^{{2 \over 3}}} - {8^{ - 1{1 \over 3}}}\) \(+ {\left( {{9^0}} \right)^2};\)
c) \({27^{{2 \over 3}}} + {\left( {{1 \over {16}}} \right)^{ - 0,75}} - {25^{0,5}}\)
d) \({\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left({2{1 \over 4}} \right)^{ - 1{1 \over 2}}} \) \(+ 19{\left( { - 3} \right)^{ - 3}}\)
Lời giải chi tiết:
\({81^{ - 0,75}} + {\left( {{1 \over {125}}} \right)^{{{ - 1} \over 3}}} - {\left({{1 \over {32}}} \right)^{{{ - 3} \over 5}}} \)
\(= {\left( {{3^4}} \right)^{ {{ - 3} \over 4}}} + {\left({{{\left( {{1 \over 5}} \right)}^3}} \right)^{{{ - 1} \over 3}}} - {\left({{{\left( {{1 \over 2}} \right)}^5}} \right)^{{{ - 3} \over 5}}}\)
\(= {\left( 3 \right)^{ - 3}} + {\left({{1 \over 5}} \right)^{ - 1}} - {\left({{1 \over 2}} \right)^{ - 3}}\)
\(= {1 \over {27}} + 5 - 8 = {1 \over {27}} - 3 = - {{80} \over {27}}\)
Cách khác:
$\begin{array}{l}
{81^{ - 0,75}} + {\left( {\frac{1}{{125}}} \right)^{\frac{{ - 1}}{3}}} - {\left( {\frac{1}{{32}}} \right)^{\frac{{ - 3}}{5}}}\\
= {81^{\frac{{ - 3}}{4}}} + {125^{\frac{1}{3}}} - {32^{\frac{3}{5}}}\\
= \frac{1}{{\sqrt[4]{{{{81}^3}}}}} + \sqrt[3]{{125}} - \sqrt[5]{{{{32}^3}}}\\
= \frac{1}{{\sqrt[4]{{{{81}^3}}}}} + 5 - {\left( {\sqrt[5]{{32}}} \right)^3}\\
= \frac{1}{{27}} + 5 - 8 = \frac{{ - 80}}{{27}}
\end{array}$
Lời giải chi tiết:
\(0,{001^{{{ - 1} \over 3}}} - {\left( { - 2} \right)^{ - 2}}{. 64^{{2 \over 3}}} - {8^{ - 1{1 \over 3}}} + {\left({{9^0}} \right)^2} \)
\(= {\left( {{{10}^{ - 3}}} \right)^{ - {1 \over 3}}} - {2^{ - 2}}.{\left({{2^6}} \right)^{{2 \over 3}}} - {\left({{2^3}} \right)^{ - {4 \over 3}}} + 1\)
\(= 10 - {2^2} - {2^{ - 4}} + 1 = 7 - {1 \over {16}} = {{111} \over {16}}\)
Cách khác:
$\begin{array}{l}
{\left( {0,001} \right)^{ - \frac{1}{3}}} - {\left( { - 2} \right)^{ - 2}}{.64^{\frac{2}{3}}} - {8^{ - 1\frac{1}{3}}} + {\left( {{9^0}} \right)^2}\\
= \left( {\frac{1}{{\sqrt[3]{{0,001}}}}} \right) - {\left( {\frac{1}{{ - 2}}} \right)^2}\sqrt[3]{{{{64}^2}}} - {8^{ - \frac{4}{3}}} + {1^2}\\
= 10 - \frac{1}{4}{\left( {\sqrt[3]{{64}}} \right)^2} - \frac{1}{{\sqrt[3]{{{8^4}}}}} + 1\\
= 10 - \frac{1}{4}.16 - \frac{1}{{16}} + 1 = \frac{{111}}{{16}}
\end{array}$
Lời giải chi tiết:
\({27^{{2 \over 3}}} + {\left( {{1 \over {16}}} \right)^{ - 0,75}} - {25^{0,5}} \)
\(= {\left( {{3^3}} \right)^{{2 \over 3}}} + {\left({{2^{ - 4}}} \right)^{ - {3 \over 4}}} - {\left({{5^2}} \right)^{{1 \over 2}}} \)
\(= {3^2} + {2^3} - 5 = 12\)
Cách khác:
$\begin{array}{l}
{27^{\frac{2}{3}}} + {\left( {\frac{1}{{16}}} \right)^{ - 0,75}} - {25^{0,5}}\\
= {\left( {\sqrt[3]{{27}}} \right)^2} + {16^{\frac{3}{4}}} - {25^{\frac{1}{2}}}\\
= {3^2} + {2^3} - 5 = 12
\end{array}$
Lời giải chi tiết:
\({\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left({2{1 \over 4}} \right)^{ - 1{1 \over 2}}} \) \(+ 19{\left( { - 3} \right)^{ - 3}} \)
\(= {\left( {{{\left( { - 2} \right)}^{ - 1}}} \right)^{ - 4}} - {\left({{5^4}} \right)^{{1 \over 4}}} - {\left({{{\left( {{3 \over 2}} \right)}^2}} \right)^{ - {3 \over 2}}} \) \(+ {{19} \over { - 27}}\)
\(= {2^4} - 5 - {\left( {{3 \over 2}} \right)^{ - 3}} - {{19} \over {27}} \)
\(= 11 - {8 \over {27}} - {{19} \over {27}} = 10.\)
Cách khác:
$\begin{array}{l}
{\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left( {2\frac{1}{4}} \right)^{ - 1\frac{1}{2}}} + 19{\left( { - 3} \right)^{ - 3}}\\
= \frac{1}{{{{\left( { - 0,5} \right)}^4}}} - \sqrt[4]{{625}} - {\left( {\frac{4}{9}} \right)^{\frac{3}{2}}} + 19.\frac{1}{{ - 27}}\\
= 16 - 5 - \frac{8}{{27}} - \frac{{19}}{{27}} = 10
\end{array}$
a) \({81^{ - 0,75}} + {\left( {{1 \over {125}}} \right)^{{{ - 1} \over 3}}} - {\left({{1 \over {32}}} \right)^{{{ - 3} \over 5}}};\)
b) \(0,{001^{{{ - 1} \over 3}}} - {\left( { - 2} \right)^{ - 2}}{. 64^{{2 \over 3}}} - {8^{ - 1{1 \over 3}}}\) \(+ {\left( {{9^0}} \right)^2};\)
c) \({27^{{2 \over 3}}} + {\left( {{1 \over {16}}} \right)^{ - 0,75}} - {25^{0,5}}\)
d) \({\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left({2{1 \over 4}} \right)^{ - 1{1 \over 2}}} \) \(+ 19{\left( { - 3} \right)^{ - 3}}\)
Câu a
\({81^{ - 0,75}} + {\left( {{1 \over {125}}} \right)^{{{ - 1} \over 3}}} - {\left({{1 \over {32}}} \right)^{{{ - 3} \over 5}}};\)Lời giải chi tiết:
\({81^{ - 0,75}} + {\left( {{1 \over {125}}} \right)^{{{ - 1} \over 3}}} - {\left({{1 \over {32}}} \right)^{{{ - 3} \over 5}}} \)
\(= {\left( {{3^4}} \right)^{ {{ - 3} \over 4}}} + {\left({{{\left( {{1 \over 5}} \right)}^3}} \right)^{{{ - 1} \over 3}}} - {\left({{{\left( {{1 \over 2}} \right)}^5}} \right)^{{{ - 3} \over 5}}}\)
\(= {\left( 3 \right)^{ - 3}} + {\left({{1 \over 5}} \right)^{ - 1}} - {\left({{1 \over 2}} \right)^{ - 3}}\)
\(= {1 \over {27}} + 5 - 8 = {1 \over {27}} - 3 = - {{80} \over {27}}\)
Cách khác:
$\begin{array}{l}
{81^{ - 0,75}} + {\left( {\frac{1}{{125}}} \right)^{\frac{{ - 1}}{3}}} - {\left( {\frac{1}{{32}}} \right)^{\frac{{ - 3}}{5}}}\\
= {81^{\frac{{ - 3}}{4}}} + {125^{\frac{1}{3}}} - {32^{\frac{3}{5}}}\\
= \frac{1}{{\sqrt[4]{{{{81}^3}}}}} + \sqrt[3]{{125}} - \sqrt[5]{{{{32}^3}}}\\
= \frac{1}{{\sqrt[4]{{{{81}^3}}}}} + 5 - {\left( {\sqrt[5]{{32}}} \right)^3}\\
= \frac{1}{{27}} + 5 - 8 = \frac{{ - 80}}{{27}}
\end{array}$
Câu b
\(0,{001^{{{ - 1} \over 3}}} - {\left( { - 2} \right)^{ - 2}}{. 64^{{2 \over 3}}} - {8^{ - 1{1 \over 3}}}\) \(+ {\left( {{9^0}} \right)^2};\)Lời giải chi tiết:
\(0,{001^{{{ - 1} \over 3}}} - {\left( { - 2} \right)^{ - 2}}{. 64^{{2 \over 3}}} - {8^{ - 1{1 \over 3}}} + {\left({{9^0}} \right)^2} \)
\(= {\left( {{{10}^{ - 3}}} \right)^{ - {1 \over 3}}} - {2^{ - 2}}.{\left({{2^6}} \right)^{{2 \over 3}}} - {\left({{2^3}} \right)^{ - {4 \over 3}}} + 1\)
\(= 10 - {2^2} - {2^{ - 4}} + 1 = 7 - {1 \over {16}} = {{111} \over {16}}\)
Cách khác:
$\begin{array}{l}
{\left( {0,001} \right)^{ - \frac{1}{3}}} - {\left( { - 2} \right)^{ - 2}}{.64^{\frac{2}{3}}} - {8^{ - 1\frac{1}{3}}} + {\left( {{9^0}} \right)^2}\\
= \left( {\frac{1}{{\sqrt[3]{{0,001}}}}} \right) - {\left( {\frac{1}{{ - 2}}} \right)^2}\sqrt[3]{{{{64}^2}}} - {8^{ - \frac{4}{3}}} + {1^2}\\
= 10 - \frac{1}{4}{\left( {\sqrt[3]{{64}}} \right)^2} - \frac{1}{{\sqrt[3]{{{8^4}}}}} + 1\\
= 10 - \frac{1}{4}.16 - \frac{1}{{16}} + 1 = \frac{{111}}{{16}}
\end{array}$
Câu c
\({27^{{2 \over 3}}} + {\left( {{1 \over {16}}} \right)^{ - 0,75}} - {25^{0,5}}\)Lời giải chi tiết:
\({27^{{2 \over 3}}} + {\left( {{1 \over {16}}} \right)^{ - 0,75}} - {25^{0,5}} \)
\(= {\left( {{3^3}} \right)^{{2 \over 3}}} + {\left({{2^{ - 4}}} \right)^{ - {3 \over 4}}} - {\left({{5^2}} \right)^{{1 \over 2}}} \)
\(= {3^2} + {2^3} - 5 = 12\)
Cách khác:
$\begin{array}{l}
{27^{\frac{2}{3}}} + {\left( {\frac{1}{{16}}} \right)^{ - 0,75}} - {25^{0,5}}\\
= {\left( {\sqrt[3]{{27}}} \right)^2} + {16^{\frac{3}{4}}} - {25^{\frac{1}{2}}}\\
= {3^2} + {2^3} - 5 = 12
\end{array}$
Câu d
\({\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left({2{1 \over 4}} \right)^{ - 1{1 \over 2}}} \) \(+ 19{\left( { - 3} \right)^{ - 3}}\)Lời giải chi tiết:
\({\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left({2{1 \over 4}} \right)^{ - 1{1 \over 2}}} \) \(+ 19{\left( { - 3} \right)^{ - 3}} \)
\(= {\left( {{{\left( { - 2} \right)}^{ - 1}}} \right)^{ - 4}} - {\left({{5^4}} \right)^{{1 \over 4}}} - {\left({{{\left( {{3 \over 2}} \right)}^2}} \right)^{ - {3 \over 2}}} \) \(+ {{19} \over { - 27}}\)
\(= {2^4} - 5 - {\left( {{3 \over 2}} \right)^{ - 3}} - {{19} \over {27}} \)
\(= 11 - {8 \over {27}} - {{19} \over {27}} = 10.\)
Cách khác:
$\begin{array}{l}
{\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left( {2\frac{1}{4}} \right)^{ - 1\frac{1}{2}}} + 19{\left( { - 3} \right)^{ - 3}}\\
= \frac{1}{{{{\left( { - 0,5} \right)}^4}}} - \sqrt[4]{{625}} - {\left( {\frac{4}{9}} \right)^{\frac{3}{2}}} + 19.\frac{1}{{ - 27}}\\
= 16 - 5 - \frac{8}{{27}} - \frac{{19}}{{27}} = 10
\end{array}$
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