Question

Tìm ĐKXĐ rồi rút gon:

\(B=\dfrac{\sqrt{x^3}}{\sqrt{xy}-2y}-\dfrac{2x}{x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}}.\dfrac{1-x}{1-\sqrt{x}}\)

Mọi người giúp em với

Answer 1

Lời giải:
ĐK: \(x\geq 0; x\neq 1; x\neq 4y; y>0\)

\(B=\frac{\sqrt{x^3}}{\sqrt{y}(\sqrt{x}-2\sqrt{y})}-\frac{2x}{(x-2\sqrt{xy})+(\sqrt{x}-2\sqrt{y})}.\frac{(1-\sqrt{x})(1+\sqrt{x})}{1-\sqrt{x}}\)

\(=\frac{\sqrt{x^3}}{\sqrt{y}(\sqrt{x}-2\sqrt{y})}-\frac{2x}{(\sqrt{x}-2\sqrt{y})(\sqrt{x}+1)}.(1+\sqrt{x})\)

\(=\frac{\sqrt{x^3}}{\sqrt{y}(\sqrt{x}-2\sqrt{y})}-\frac{2x}{\sqrt{x}-2\sqrt{y}}\)

\(=\frac{\sqrt{x^3}-2x\sqrt{y}}{\sqrt{y}(\sqrt{x}-2\sqrt{y})}\)

\(=\frac{x(\sqrt{x}-2\sqrt{y})}{\sqrt{y}(\sqrt{x}-2\sqrt{y})}=\frac{x}{\sqrt{y}}\)