Question

Cho A= \(\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+2}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{2\sqrt{z}}{\sqrt{zx}+2\sqrt{z}+2}\)

Biết xyz=4. Tính \(\sqrt{A}\)

Answer 1

Lời giải:

Đặt $(\sqrt{x}, \sqrt{y}, \sqrt{z})=(a,b,c)$. Khi đó:

$abc=\sqrt{xyz}=2$

$A=\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ca+2c+2}$

$=\frac{a}{ab+a+2}+\frac{ab}{abc+ab+a}+\frac{2c}{ca+2c+abc}$

$=\frac{a}{ab+a+2}+\frac{ab}{2+ab+a}+\frac{2}{a+2+ab}$

$=\frac{a+ab+2}{ab+a+2}=1$

$\Rightarrow \sqrt{A}=1$

Vậy.........

Answer 2

Ta có: \(xyz=4\Leftrightarrow\sqrt{xyz}=\sqrt{4}=2\)

\( A=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+2}+\frac{\sqrt{y}}{\sqrt{xy}+\sqrt{y}+1}+\frac{2\sqrt{z}}{\sqrt{zx}+2\sqrt{z}+2}\)

\(=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\frac{\sqrt{xy}}{\sqrt{xyz}+\sqrt{xy}+\sqrt{x}}+\frac{2\sqrt{z}}{\sqrt{xz}+\sqrt{xyz}\sqrt{z}+\sqrt{xyz}}\\ =\frac{\sqrt{x}}{\sqrt{xyz}+\sqrt{xy}+\sqrt{x}}+\frac{\sqrt{xy}}{\sqrt{xyz}+\sqrt{xy}+\sqrt{x}}+\frac{2}{\sqrt{xyz}+\sqrt{xy}+\sqrt{x}}\\ =\frac{\sqrt{x}+\sqrt{xy}+2}{\sqrt{xyz}+\sqrt{xy}+\sqrt{z}}\\ =\frac{\sqrt{x}+\sqrt{xy}+\sqrt{xyz}}{\sqrt{xyz}+\sqrt{xy}+\sqrt{x}}\\ =1\)

\(\Leftrightarrow A=1\\ \Rightarrow\sqrt{A}=\sqrt{1}=1\)