Bài 2.2 trang 99 SBT giải tích 12

Tính:

a) \(27^{\dfrac{2}{3}} - (-2)^{-2} +(3\dfrac{3}{8})^{-\dfrac{1}{3}} \)

b) \( ( - 0.5)^{{-4}} - 625^{0.25} - (2\dfrac{1}{4})^{-1\dfrac{1}{2}} \)

Lời giải

a) \( 27^{\dfrac{2}{3}} - \Big(-2\Big)^{-2} +\Big(3\dfrac{3}{8}\Big)^{-\dfrac{1}{3}} \)

\( = \Big(3^3\Big)^{\dfrac{2}{3}} - \dfrac{1}{4} + \Big(\dfrac{27}{8}\Big)^{-\dfrac{1}{3}} = 3^{3.\dfrac{2}{3}} - \dfrac{1}{4} + \Big(\dfrac{3}{2} \Big)^{{3}^{-\dfrac{1}{3}}} \)

\( = 3^2 - \dfrac{1}{4} + \Big(\dfrac{3}{2}\Big)^{-1}\)

\(=9 - \dfrac{1}{4} + \dfrac{2}{3}\)

\(= \dfrac{113}{12} \)

b)\( ( - 0.5)^{{-4}} - 625^{0.25} - \Big(2\dfrac{1}{4}\Big)^{-1\dfrac{1}{2}} \)

\(=\Big( \dfrac{-1}{2}\Big)^{-4} - \Big( 5^{4}\Big)^{0.25} - \Big(\dfrac{9}{4}\Big)^{\dfrac{-3}{2}}\)

\(= 16 - 5^1 - {\Big( \Big( \dfrac{3}{2}\Big)^{2}\Big)}^{-\dfrac{3}{2}}\)

\(= 16 - 5 - \Big( \dfrac{3}{2}\Big)^{-3}\)

\(= 11 - \dfrac{8}{27} =\dfrac{289}{27}\)