Bài 1 trang 33 SGK Toán 11 tập 2 – Cánh Diều

Đề bài

Tính:

a)     \({\left( {\frac{1}{{256}}} \right)^{ – 0,75}} + {\left( {\frac{1}{{27}}} \right)^{ – \frac{4}{3}}}\)

b)    \({\left( {\frac{1}{{49}}} \right)^{ – 1,5}} – {\left( {\frac{1}{{125}}} \right)^{ – \frac{2}{3}}}\)

c)     \(\left( {{4^{3 + \sqrt 3 }} – {4^{\sqrt 3  – 1}}} \right){.2^{ – 2\sqrt 3 }}\)

Lời giải chi tiết

a)     \({\left( {\frac{1}{{256}}} \right)^{ – 0,75}} + {\left( {\frac{1}{{27}}} \right)^{ – \frac{4}{3}}} = {\left( {\frac{1}{{256}}} \right)^{ – \frac{3}{4}}} + {\left( {\frac{1}{{27}}} \right)^{ – \frac{4}{3}}} = {256^{\frac{3}{4}}} + {27^{\frac{4}{3}}} = \sqrt[4]{{{{256}^3}}} + \sqrt[3]{{{{27}^4}}} = \sqrt[4]{{{{\left( {{2^8}} \right)}^3}}} + \sqrt[3]{{{{\left( {{3^3}} \right)}^4}}}\)

\( = \sqrt[4]{{{2^{24}}}} + \sqrt[3]{{{3^{12}}}} = {2^6} + {3^4} = 145\)

b)    

${{\left( \frac{1}{49} \right)}^{-1,5}}-{{\left( \frac{1}{125} \right)}^{-\frac{2}{3}}}={{\left( \frac{1}{49} \right)}^{\frac{3}{2}}}-{{\left( \frac{1}{125} \right)}^{\frac{2}{3}}}={{\left( \frac{1}{7} \right)}^{2.\frac{-3}{2}}}\text{ }\!\!~\!\!\text{ }-{{\left( \frac{1}{5} \right)}^{3.\frac{-2}{3}}}$

$={{\left( {{7}^{-1}} \right)}^{-3}}-{{\left( {{5}^{-1}} \right)}^{-2}}={{7}^{3}}-{{5}^{2}}=318$

c)     \(\left( {{4^{3 + \sqrt 3 }} – {4^{\sqrt 3  – 1}}} \right){.2^{ – 2\sqrt 3 }} = \left( {{2^{6 + 2\sqrt 3 }} – {2^{2\sqrt 3  – 2}}} \right){.2^{ – 2\sqrt 3 }} = {2^6} – {2^{ – 2}} = 64 – \frac{1}{4} = \frac{{255}}{4}\)