Question

Giúp e bài toán này ak

Answer 1

ĐKXĐ: \(x\ge0;y\ge0;x+y\ne0\)

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

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

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

Ta có \(x+y-\sqrt{xy}=x-\sqrt{xy}+\frac{y}{4}+\frac{3y}{4}=\left(\sqrt{x}-\frac{\sqrt{y}}{2}\right)^2+\frac{3y}{4}>0\)

\(\Rightarrow P=\frac{\sqrt{xy}}{x+y-\sqrt{xy}}\ge0\)

Với \(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\) \(\Rightarrow B=0=\sqrt{B}\)

Với \(xy\ne0\), chia cả tử và mẫu cho \(\sqrt{xy}\)

\(P=\frac{1}{\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}-1}\le\frac{1}{2\sqrt{\sqrt{\frac{xy}{yx}}}-1}=1\)

\(\Rightarrow0< P\le1\Rightarrow\sqrt{P}\ge P\)

Answer 2