Question

cho A = \(\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\) đk x > 0, x khác 4

a, rút gọn A

b, tìm x đề A = -1

Answer 1

a.

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

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

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

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

b.

\(A=-1\Leftrightarrow\frac{4x}{\sqrt{x}-3}=-1\)

\(\Leftrightarrow4x=-\sqrt{x}+3\)

\(\Leftrightarrow4x+\sqrt{x}-3=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=\frac{3}{4}\end{matrix}\right.\)

Vậy \(A=-1\Leftrightarrow x=\frac{3}{4}\)