Question

Cho biểu thức :

A = \(\left(\frac{4\sqrt{x}}{2+\sqrt{x}}-\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\). ĐKXĐ : x > 0, x ≠ 4

a) Rút gọn A

b) Tìm x để A = -1

Answer 1

Lời giải:

a)

\(A=\left[\frac{4\sqrt{x}(\sqrt{x}-2)}{(2+\sqrt{x})(\sqrt{x}-2)}-\frac{8x}{(\sqrt{x}-2)(\sqrt{x}+2)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-2)}-\frac{2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\right]\)

\(=\frac{4\sqrt{x}(\sqrt{x}-2)-8x}{(\sqrt{x}-2)(\sqrt{x}+2)}:\frac{\sqrt{x}-1-2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\)

\(=\frac{-4x-8\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}(\sqrt{x}-2)}{3-\sqrt{x}}\)

\(=\frac{-4\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}(\sqrt{x}-2)}{3-\sqrt{x}}=\frac{4x}{\sqrt{x}-3}\)

b)

Để $A=-1\Leftrightarrow \frac{4x}{\sqrt{x}-3}=-1$

$\Leftrightarrow 4x+\sqrt{x}-3=0$

$\Leftrightarrow (4\sqrt{x}-3)(\sqrt{x}+1)=0$

$\Rightarrow 4\sqrt{x}-3=0$

$\Leftrightarrow x=\frac{9}{16}$ (thỏa mãn ĐKXĐ)

Vậy........