Question

Сократите это жду решений

Answer 1

   $(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab})*(\frac{\sqrt{a}+\sqrt{b}}{a-b})^2=$

$=(\frac{a\sqrt{a}+b\sqrt{b}-(\sqrt{a}+\sqrt{b})*\sqrt{ab} }{\sqrt{a}+\sqrt{b}})*(\frac{\sqrt{a}+\sqrt{b}}{a-b})^2=$

$=(\frac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{a} }{\sqrt{a}+\sqrt{b}})*(\frac{\sqrt{a}+\sqrt{b}}{a-b})^2=$

$=\frac{(a\sqrt{a}-b\sqrt{a})-(a\sqrt{b}-b\sqrt{b}) }{\sqrt{a}+\sqrt{b}}*(\frac{\sqrt{a}+\sqrt{b}}{a-b})^2=$

$=\frac{\sqrt{a}(a-b)-\sqrt{b}(a-b)}{\sqrt{a}+\sqrt{b}}*(\frac{\sqrt{a}+\sqrt{b}}{a-b})^2=$

$=\frac{(\sqrt{a}-\sqrt{b})(a-b)}{\sqrt{a}+\sqrt{b}}*\frac{(\sqrt{a}+\sqrt{b})^2}{(a-b)^2}=$

$=\frac{(\sqrt{a}-\sqrt{b})(a-b)*(\sqrt{a}+\sqrt{b})*(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})*(a-b)*(a-b)}=$

$=\frac{(\sqrt{a}-\sqrt{b})*(\sqrt{a}+\sqrt{b})}{a-b}=\frac{(\sqrt{a})^2-(\sqrt{b})^2}{a-b}=$

$=\frac{a-b}{a-b}=1$