Question

\(A=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+1}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)

Rút gọn A.Tìm GTLN của A

 

Answer 1

Lời giải:

ĐK: $x\geq 0; x\neq 1$

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

\(=\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{2(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{2}=\frac{(\sqrt{x}-2)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}+1)}{2(\sqrt{x}+1)}=\frac{-6\sqrt{x}}{2(\sqrt{x}+2)}=\frac{-3\sqrt{x}}{\sqrt{x}+2}\)

Vì $x\geq 0$ nên $3\sqrt{x}\geq 0; \sqrt{x}+2>0$

$\Rightarrow \frac{3\sqrt{x}}{\sqrt{x}+2}\geq 0$

$\Rightarrow A\leq 0$ hay $A_{\max}=0$ khi $x=0$