Question

П о м о г и т е!!!!! п о ж а л у й с т а ​

Answer 1

Объяснение:

$\tt\displaystyle \[{\left( {\left( {{a^{-\frac{3}{2}}}b}\right){{\left( {a{b^{-2}}} \right)}^{-\frac{1}{2}}}{{\left( {{a^{-1}}} \right)}^{-\frac{2}{3}}}} \right)^3}={\left( {\left( {{a^{-\frac{3}{2}}}b} \right){{\left( {{a^{-\frac{1}{2}}}{b^{-2*\left( {-\frac{1}{2}} \right)}}} \right)}^{}}{{\left( {{a^{-1*\left( {-\frac{2}{3}} \right)}}} \right)}^{}}} \right)^3}=\]$

$\tt\displaystyle \[={\left( {\left( {{a^{-\frac{3}{2}}}b} \right){{\left( {{a^{-\frac{1}{2}}}{b^1}} \right)}^{}}{{\left( {{a^{\frac{2}{3}}}} \right)}^{}}} \right)^3}={\left( {{a^{-\frac{3}{2}+\left( {-\frac{1}{2}} \right)+\frac{2}{3}}}{b^{1+1}}}\right)^3}={\left( {{a^{-\frac{4}{3}}}{b^2}} \right)^3}=\]$

$= {a^{-\frac{4}{3}*3}}{b^{2*3}}={a^{-4}}{b^6}=\boxed{\frac{{{b^6}}}{{{a^4}}}}$