Bài 3 trang 33 SGK Toán 11 tập 2 - Cánh Diều

Đề bài

Rút gọn mỗi biểu thức sau:

a)     \(\frac{{{a^{\frac{7}{3}}} - {a^{\frac{1}{3}}}}}{{{a^{\frac{4}{3}}} - {a^{\frac{1}{3}}}}} - \frac{{{a^{\frac{5}{3}}} - {a^{ - \frac{1}{3}}}}}{{{a^{\frac{2}{3}}} + {a^{ - \frac{1}{3}}}}}\,\,\,(a > 0;a \ne 1)\)

b)    \(\frac{{{{\left( {\sqrt[4]{{{a^3}{b^2}}}} \right)}^4}}}{{\sqrt[4]{{\sqrt {{a^{12}}{b^6}} }}}}\,\,\,(a > 0;b > 0)\)

Lời giải chi tiết

a)     \(\frac{{{a^{\frac{7}{3}}} - {a^{\frac{1}{3}}}}}{{{a^{\frac{4}{3}}} - {a^{\frac{1}{3}}}}} - \frac{{{a^{\frac{5}{3}}} - {a^{ - \frac{1}{3}}}}}{{{a^{\frac{2}{3}}} + {a^{ - \frac{1}{3}}}}} = \left( {{a^{\frac{7}{3}}} - {a^{\frac{1}{3}}}} \right):\left( {{a^{\frac{4}{3}}} - {a^{\frac{1}{3}}}} \right) - \left( {{a^{\frac{5}{3}}} - {a^{ - \frac{1}{3}}}} \right):\left( {{a^{\frac{2}{3}}} + {a^{ - \frac{1}{3}}}} \right)\)

\( = {a^{\frac{7}{3}}}:{a^{\frac{4}{3}}} - {a^{\frac{7}{3}}}:{a^{\frac{1}{3}}} - {a^{\frac{1}{3}}}:{a^{\frac{4}{3}}} - {a^{\frac{5}{3}}}:{a^{\frac{2}{3}}} - {a^{\frac{5}{3}}}:{a^{ - \frac{1}{3}}} + {a^{ - \frac{1}{3}}}:{a^{\frac{2}{3}}} + {a^{ - \frac{1}{3}}}:{a^{ - \frac{1}{3}}}\)

\( = {a^1} - {a^2} - {a^{ - 1}} - {a^1} - {a^2} + {a^{ - 1}} + {a^0} = {a^0} = 1\)

b)    \(\frac{{{{\left( {\sqrt[4]{{{a^3}{b^2}}}} \right)}^4}}}{{\sqrt[4]{{\sqrt {{a^{12}}{b^6}} }}}} = \frac{{{a^3}{b^2}}}{{\sqrt[4]{{\sqrt {{{\left( {{a^6}{b^3}} \right)}^2}} }}}} = \frac{{{a^3}{b^2}}}{{\sqrt[4]{{{a^6}{b^3}}}}} = {a^3}{b^2}:{a^{\frac{6}{4}}}{b^{\frac{3}{4}}} = {a^{\frac{3}{2}}}{b^{\frac{5}{4}}} = \sqrt {{a^3}} .\sqrt[4]{{{b^5}}}\)