Question

[Đào Ngọc Quý]

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

a) Rút gọn B

b) Chứng minh: \(B\ge0\)

c) So sánh B với \(\sqrt{B}\)

Answer 1

[Akai Haruma]

Lời giải:

ĐK: $x\neq y; x,y\geq 0$

a)

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

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

\(=\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}.\frac{1}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}(\sqrt{x}-\sqrt{y})}{\sqrt{x}-\sqrt{y}}.\frac{1}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

b) Ta thấy:

\(\sqrt{xy}\geq 0, \forall x,y\geq 0\)

\(x-\sqrt{xy}+y=(\sqrt{x}-\frac{\sqrt{y}}{2})^2+\frac{3}{4}y>0, \forall x,y\geq 0; x\neq y\)

\(\Rightarrow B=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\geq 0\) (đpcm)

c)

Áp dụng BĐT AM-GM: \(x+y\geq 2\sqrt{xy}\Rightarrow x-\sqrt{xy}+y\geq \sqrt{xy}\)

\(\Rightarrow B=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\leq 1\)

Dấu "=" xảy ra khi $x=y$. Mà $x\neq y$ nên $B< 1\Rightarrow \sqrt{B}< 1$

Do đó: \(B=\sqrt{B}.\sqrt{B}< \sqrt{B}\)

Answer 2

[Đào Ngọc Quý]

bạn ơi ở phần c áp dụng cái j vậy bạn?