Toán 12 Lũy thừa, số mũ

a) $\dfrac{ \left ( \sqrt[4]{a^3b^2} \right )^4}{ \sqrt[3]{ \sqrt{a^{12}b^6}}}$ $(DKXD: \ a, b > 0)$
$= \dfrac{ \left [ \left ( a^3b^2 \right )^4 \right ]^{1/4}}{ \left [ \left ( a^{12}b^6 \right )^{1/2} \right ]^{1/3}} \\
\\
= \dfrac{ a^3b^2 }{ \left ( a^{12}b^6 \right )^{1/6} } \\
\\
= \dfrac{ a^3b^2 }{ a^2b } \\
= ab
$

c) $ \left ( \dfrac{a^{ \sqrt{3}}}{b^{ \sqrt{3} - 1}} \right )^{ \sqrt{3} +1}. \dfrac{a^{ -1- \sqrt{3}}}{b^{ -2}}$
$= \dfrac{a^{ \sqrt{3}.( \sqrt{3}+1)}}{b^{( \sqrt{3} - 1)( \sqrt{3}+1)}}. \left ( a^{ -1- \sqrt{3}}.b^2 \right ) \\
\\
= \dfrac{a^{ 3+ \sqrt{3}}}{b^{3-1}}. \left ( a^{ -1- \sqrt{3}}.b^2 \right ) \\
\\
= \dfrac{a^{ 3+ \sqrt{3}}}{b^2}. \left ( a^{ -1- \sqrt{3}}.b^2 \right ) \\
= \left ( a^{ 3+ \sqrt{3}} \right ). \left ( a^{ -1- \sqrt{3}} \right ) \\
= a^{ 3+ \sqrt{3} -1- \sqrt{3}} \\
= a^2$