Bài 6.2 trang 9 SGK Toán 11 tập 2 - Kết nối tri thức

a) \({27^{\frac{2}{3}}} + {81^{ - 0,75}} - {25^{0,5}}\)

\(= \sqrt[3]{{{{27}^2}}} + {81^{ - \frac{3}{4}}} - {25^{\frac{1}{2}}}\)

\(= {\left( {\sqrt[3]{{{3^3}}}} \right)^2} + \frac{1}{{\sqrt[4]{{{{81}^3}}}}} - \sqrt {25} \)

\(= {3^2} + \frac{1}{{{{\left( {\sqrt[4]{{{3^4}}}} \right)}^3}}} - 5\)

\(= 9 + \frac{1}{{{3^3}}} - 5 \)

\(= 9 + \frac{1}{{27}} - 5 \)

\(= \frac{{109}}{{27}}\).

b) \({4^{2 - 3\sqrt 7 }}{.8^{2\sqrt 7 }} \)

\(= {\left( {{2^2}} \right)^{2 - 3\sqrt 7 }}.{\left( {{2^3}} \right)^{2\sqrt 7 }}\)

\(= {2^{2.\left( {2 - 3\sqrt 7 } \right)}}{.2^{3.2\sqrt 7 }}\)

\(= {2^{4 - 6\sqrt 7 }}{.2^{6\sqrt 7 }} \)

\(= {2^{4 - 6\sqrt 7  + 6\sqrt 7 }} \)

\(= {2^4} \)

\(= 16\).