Bài tập 4 trang 76 SGK Toán 12 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{81^{ - 0,75}} + {\left( {\frac{1}{{125}}} \right)^{\frac{{ - 1}}{3}}} - {\left( {\frac{1}{{32}}} \right)^{\frac{{ - 3}}{5}}}\\
 = {({3^4})^{\frac{{ - 3}}{4}}} + {\left( {{{\left( {\frac{1}{5}} \right)}^3}} \right)^{ - \frac{1}{3}}} - {\left( {{{\left( {\frac{1}{2}} \right)}^5}} \right)^{\frac{{ - 3}}{5}}}
\end{array}\\
\begin{array}{l}
 = {(3)^{ - 3}} + {\left( {\frac{1}{5}} \right)^{ - 1}} - \left( {{{\frac{1}{2}}^{ - 3}}} \right)\\
 = \frac{1}{{27}} + 5 - 8 = \frac{1}{{27}} - 3 = \frac{{ - 80}}{{27}}
\end{array}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
0,{001^{\frac{{ - 1}}{3}}} - {\left( { - 2} \right)^{ - 2}}{.64^{\frac{2}{3}}} - {8^{ - 1\frac{1}{3}}} + {\left( {{9^0}} \right)^2}\\
 = {\left( {{{10}^{ - 3}}} \right)^{ - \frac{1}{3}}} - {2^{ - 2}}.{\left( {{2^6}} \right)^{\frac{2}{3}}} - {\left( {{2^3}} \right)^{ - \frac{4}{3}}} + 1
\end{array}\\
{ = 10 - {2^2} - {2^{ - 4}} + 1 = 7 - \frac{1}{{16}} = \frac{{111}}{{16}}}
\end{array}\)

c)

\(\begin{array}{l}
{27^{\frac{2}{3}}} + {\left( {\frac{1}{{16}}} \right)^{ - 0,75}} - {25^{0,5}}\\
 = {({3^3})^{\frac{2}{3}}} + {({2^{ - 4}})^{ - \frac{3}{4}}} - {({5^2})^{\frac{1}{2}}}\\
 = {3^2} + {2^3} - 5 = 12
\end{array}\)

d)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{( - 0,5)^{ - 4}} - {625^{0,25}} - {\left( {2\frac{1}{4}} \right)^{ - 1\frac{1}{2}}} + 19.{\left( { - 3} \right)^{ - 3}}\\
 = {\left( {{{\left( { - 2} \right)}^{ - 1}}} \right)^{ - 4}} - {\left( {{5^4}} \right)^{\frac{1}{4}}} - {\left( {{{\left( {\frac{3}{2}} \right)}^2}} \right)^{ - \frac{3}{2}}} + \frac{{19}}{{ - 27}}
\end{array}\\
{ = {2^4} - 5 - {{\left( {\frac{3}{2}} \right)}^{ - 3}} - \frac{{19}}{{27}} = 11 - \frac{8}{{27}} - \frac{{19}}{{27}} = 10.}
\end{array}\)

-- Mod Toán 12 HỌC247