Bài tập 2.2 trang 99 SBT Toán 12

a)

\(\begin{array}{l}
{27^{\frac{2}{3}}} - {\left( { - 2} \right)^{ - 2}} + {\left( {3\frac{3}{8}} \right)^{\frac{{ - 1}}{3}}} = {\left( {{3^3}} \right)^{\frac{2}{3}}} - \frac{1}{{{{\left( { - 2} \right)}^2}}} + {\left( {\frac{{27}}{8}} \right)^{\frac{{ - 1}}{3}}}\\
 = {3^{3.\frac{2}{3}}} - \frac{1}{4} + {\left[ {{{\left( {\frac{3}{2}} \right)}^3}} \right]^{\frac{{ - 1}}{3}}} = {3^2} - \frac{1}{4} + {\left( {\frac{3}{2}} \right)^{ - 1}}\\
 = 9 - \frac{1}{4} + \frac{2}{3} = \frac{{113}}{{12}}
\end{array}\)

b)

\(\begin{array}{l}
{\left( { - 0,5} \right)^{ - 4}} - {625^{0,25}} - {\left( {2\frac{1}{4}} \right)^{ - 1\frac{1}{2}}} = {\left( { - \frac{1}{2}} \right)^{ - 4}} - {\left( {{5^4}} \right)^{\frac{1}{4}}} - {\left( {\frac{9}{4}} \right)^{\frac{{ - 3}}{2}}}\\
 = {\left( { - 2} \right)^4} - {\left( {{5^4}} \right)^{\frac{1}{4}}} - {\left( {\frac{9}{4}} \right)^{\frac{{ - 3}}{2}}}\\
 = {( - 2)^4} - {5^{4.\frac{1}{4}}} - {\left[ {{{\left( {\frac{3}{2}} \right)}^2}} \right]^{ - \frac{3}{2}}}\\
 = 16 - 5 - {\left( {\frac{3}{2}} \right)^{ - 3}} = 11 - {\left( {\frac{2}{3}} \right)^3} = 11 - \frac{8}{{27}} = \frac{{289}}{{27}}
\end{array}\)

-- Mod Toán 12 HỌC247