Bài 18 trang 81 SGK Đại số và Giải tích 12 Nâng cao

Viết các biểu thức sau dưới dạng lũy thừa của một số với số mũ hữu tỉ:

LG a

\(\root 4 \of {{x^2}\root 3 \of x } \,\,\,\,\left( {x > 0} \right);\)       

Phương pháp giải:

Lưu ý: \(\root n \of a = a^{1 \over n} (a>0)\); \(a^m.a^n=a^{m+n}\)

Lời giải chi tiết:

\(\root 4 \of {{x^2}\root 3 \of x }  = {\left( {{x^2}.{x^{{1 \over 3}}}} \right)^{{1 \over 4}}} = {\left( {{x^{{7 \over 3}}}} \right)^{{1 \over 4}}} = {x^{{7 \over {12}}}}\)

Cách trình bày khác:

\(\begin{array}{l}
\sqrt[4]{{{x^2}\sqrt[3]{x}}} = \sqrt[4]{{{x^2}.{x^{\frac{1}{3}}}}} = \sqrt[4]{{{x^{\frac{7}{3}}}}}\\
= {\left( {{x^{\frac{7}{3}}}} \right)^{\frac{1}{4}}} = {x^{\frac{7}{3}.\frac{1}{4}}} = {x^{\frac{7}{{12}}}}
\end{array}\)

LG b

\(\root 5 \of {{b \over a}\root 3 \of {{a \over b}} } \,\,\,\,\left( {a > 0,b > 0} \right);\)

Lời giải chi tiết:

\(\root 5 \of {{b \over a}\root 3 \of {{a \over b}} }  = {\left( {{b \over a}{{\left( {{a \over b}} \right)}^{{1 \over 3}}}} \right)^{{1 \over 5}}} \)

\(= {\left( {{{\left( {{a \over b}} \right)}^{ - 1}}{{\left( {{a \over b}} \right)}^{{1 \over 3}}}} \right)^{{1 \over 5}}} = {\left( {{{\left( {{a \over b}} \right)}^{ - {2 \over 3}}}} \right)^{{1 \over 5}}} \)

\(= {\left( {{a \over b}} \right)^{ - {2 \over {15}}}}\)

Cách trình bày khác:

\(\begin{array}{l}
\sqrt[5]{{\frac{b}{a}.\sqrt[3]{{\frac{a}{b}}}}} = \sqrt[5]{{{{\left( {\frac{a}{b}} \right)}^{ - 1}}.{{\left( {\frac{a}{b}} \right)}^{\frac{1}{3}}}}}\\
= \sqrt[5]{{{{\left( {\frac{a}{b}} \right)}^{ - 1}}.{{\left( {\frac{a}{b}} \right)}^{\frac{1}{3}}}}} = \sqrt[5]{{{{\left( {\frac{a}{b}} \right)}^{ - \frac{2}{3}}}}}\\
= {\left[ {{{\left( {\frac{a}{b}} \right)}^{ - \frac{2}{3}}}} \right]^{\frac{1}{5}}} = {\left( {\frac{a}{b}} \right)^{ - \frac{2}{3}.\frac{1}{5}}}\\
= {\left( {\frac{a}{b}} \right)^{ - \frac{2}{{15}}}}
\end{array}\)

LG c

\(\root 3 \of {{2 \over 3}\root 3 \of {{2 \over 3}} \sqrt {{2 \over 3}} } ;\) 

Lời giải chi tiết:

\(\begin{array}{l}
\sqrt[3]{{\frac{2}{3}\sqrt[3]{{\frac{2}{3}\sqrt {\frac{2}{3}} }}}} = \sqrt[3]{{\frac{2}{3}\sqrt[3]{{\frac{2}{3}.{{\left( {\frac{2}{3}} \right)}^{\frac{1}{2}}}}}}}\\
= \sqrt[3]{{\frac{2}{3}\sqrt[3]{{{{\left( {\frac{2}{3}} \right)}^{\frac{3}{2}}}}}}} = \sqrt[3]{{\frac{2}{3}{{\left[ {{{\left( {\frac{2}{3}} \right)}^{\frac{3}{2}}}} \right]}^{\frac{1}{3}}}}}\\
= \sqrt[3]{{\frac{2}{3}.{{\left( {\frac{2}{3}} \right)}^{\frac{3}{2}.\frac{1}{3}}}}} = \sqrt[3]{{\frac{2}{3}.{{\left( {\frac{2}{3}} \right)}^{\frac{1}{2}}}}}\\
= \sqrt[3]{{{{\left( {\frac{2}{3}} \right)}^{1 + \frac{1}{2}}}}} = \sqrt[3]{{{{\left( {\frac{2}{3}} \right)}^{\frac{3}{2}}}}}\\
= {\left[ {{{\left( {\frac{2}{3}} \right)}^{\frac{3}{2}}}} \right]^{\frac{1}{3}}} = {\left( {\frac{2}{3}} \right)^{\frac{3}{2}.\frac{1}{3}}}\\
= {\left( {\frac{2}{3}} \right)^{\frac{1}{2}}}
\end{array}\)

LG d

\(\sqrt {a\sqrt {a\sqrt {a\sqrt a } } } :{a^{{{11} \over {16}}}}\,\,\,\,\left( {a > 0} \right).\)

Lời giải chi tiết:

\(\begin{array}{l}
\sqrt {a\sqrt {a\sqrt {a\sqrt a } } } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {a\sqrt {a.{a^{\frac{1}{2}}}} } } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {a\sqrt {{a^{\frac{3}{2}}}} } } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {a{{\left( {{a^{\frac{3}{2}}}} \right)}^{\frac{1}{2}}}} } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {a.{a^{\frac{3}{2}.\frac{1}{2}}}} } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {a.{a^{\frac{3}{4}}}} } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {{a^{1 + \frac{3}{4}}}} } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {{a^{\frac{7}{4}}}} } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a.{{\left( {{a^{\frac{7}{4}}}} \right)}^{\frac{1}{2}}}} :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a.{a^{\frac{7}{4}.\frac{1}{2}}}} :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a.{a^{\frac{7}{8}}}} :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {{a^{1 + \frac{7}{8}}}} :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {{a^{\frac{{15}}{8}}}} :{a^{\frac{{11}}{{16}}}}\\
= {\left( {{a^{\frac{{15}}{8}}}} \right)^{\frac{1}{2}}}:{a^{\frac{{11}}{{16}}}}\\
= {a^{\frac{{15}}{8}.\frac{1}{2}}}:{a^{\frac{{11}}{{16}}}}\\
= {a^{\frac{{15}}{{16}}}}:{a^{\frac{{11}}{{16}}}}\\
= {a^{\frac{{15}}{{16}} - \frac{{11}}{{16}}}}\\
= {a^{\frac{1}{4}}}
\end{array}\)