- LG a
- LG b
- LG c
- LG d
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:
\[\begin{array}{l}
\sqrt {a\sqrt {a\sqrt {a\sqrt a } } } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {a\sqrt {a.{a^{\frac{1}{2}}}} } } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {a\sqrt {{a^{\frac{3}{2}}}} } } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {a{{\left[ {{a^{\frac{3}{2}}}} \right]}^{\frac{1}{2}}}} } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {a.{a^{\frac{3}{2}.\frac{1}{2}}}} } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {a.{a^{\frac{3}{4}}}} } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {{a^{1 + \frac{3}{4}}}} } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a\sqrt {{a^{\frac{7}{4}}}} } :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a.{{\left[ {{a^{\frac{7}{4}}}} \right]}^{\frac{1}{2}}}} :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a.{a^{\frac{7}{4}.\frac{1}{2}}}} :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {a.{a^{\frac{7}{8}}}} :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {{a^{1 + \frac{7}{8}}}} :{a^{\frac{{11}}{{16}}}}\\
= \sqrt {{a^{\frac{{15}}{8}}}} :{a^{\frac{{11}}{{16}}}}\\
= {\left[ {{a^{\frac{{15}}{8}}}} \right]^{\frac{1}{2}}}:{a^{\frac{{11}}{{16}}}}\\
= {a^{\frac{{15}}{8}.\frac{1}{2}}}:{a^{\frac{{11}}{{16}}}}\\
= {a^{\frac{{15}}{{16}}}}:{a^{\frac{{11}}{{16}}}}\\
= {a^{\frac{{15}}{{16}} - \frac{{11}}{{16}}}}\\
= {a^{\frac{1}{4}}}
\end{array}\]