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)    \({\left( {\frac{1}{{49}}} \right)^{ - 1,5}} - {\left( {\frac{1}{{256}}} \right)^{ - \frac{2}{3}}}\)

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)    \({\left( {\frac{1}{{49}}} \right)^{ - 1,5}} - {\left( {\frac{1}{{256}}} \right)^{ - \frac{2}{3}}} = {49^{\frac{3}{2}}} - {256^{\frac{2}{3}}} = \sqrt {{{\left( {{7^2}} \right)}^3}}  - \sqrt[3]{{{{\left( {{2^8}} \right)}^2}}} = {7^3} - \sqrt[3]{{{2^{16}}}} = {7^3} - 32\sqrt[3]{2}\)

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