Question

Giúp e với ạ.

Answer 1

\(A=\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\left(a>0;b>0;a\ne b\right)\)

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

\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

\(=a-b\)

Answer 2

\(A=\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)

\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

\(=a-b\)

Answer 3

\(A=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\cdot\dfrac{\left(\sqrt{a}-\sqrt[]{b}\right)}{1}\) =\(a-b\)   \(DK:a,b\) ≥ 0

Answer 4

\(A=\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\)

\(A=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\sqrt{a}-\sqrt{b}\)

\(A=\left(\sqrt{a}+\sqrt{b}\right).\left(\sqrt{a}-\sqrt{b}\right)\)

\(A=a-b\)