Question

[Le DuyDucc_]

Giúp mình với

Answer 1

[Nguyễn Huy Tú]

a, Với a ; b > 0 

\(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)

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

b, Với x ; y > 0 ; \(x\ne y\)

\(B=\left(\frac{x\sqrt{x}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}+\sqrt{xy}\right):\left(x-y\right)-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

\(=\left(\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\sqrt{xy}\right):\left(x-y\right)-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

\(=\frac{2\sqrt{xy}}{x-y}-\frac{2\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{2\sqrt{xy}-2\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{x-y}=\frac{-2y}{x-y}\)