Question

Cho biểu thức : P=\(\frac{\sqrt{x}+1}{x-1}\)-\(\frac{x+2}{x\sqrt{x}-1}\)-\(\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

Tìm GTLN của Q=\(\frac{2}{P}\)+\(\sqrt{x}\)

Answer 1

Lời giải:

ĐK: $x\geq 0; x\neq 1$

$P=\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}$

$=\frac{1}{\sqrt{x}-1}=-\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{x-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}$

$=\frac{x+\sqrt{x}+1-(x+2)-(x-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}$

$=\frac{-\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}=\frac{-\sqrt{x}}{x+\sqrt{x}+1}$

$\Rightarrow Q=\frac{2(x+\sqrt{x}+1)}{-\sqrt{x}}+\sqrt{x}$

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

Dễ thấy $\sqrt{x}+\frac{2}{\sqrt{x}}+2\geq 2\sqrt{2}+2$ theo BĐT Cô-si

$\Rightarrow Q\leq -(2\sqrt{2}+2)$ hay $Q_{\max}=-(2\sqrt{2}+2)$