Question

(\(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\) - \(\sqrt{ab}\) ) \(\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)

 

Answer 1

\(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)

\(=\left(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\cdot\left(\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)}\right)^2\)

\(=\dfrac{\left(a-\sqrt{ab}+b-\sqrt{ab}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

=1