\(\text{Đ}K\text{X}\text{Đ}:x\ne\pm2\)

Ta có: \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right)\div\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)

\(=\left(\frac{2x+2-4}{\left(x+2\right)^2}\right):\left(\frac{2-x-2}{\left(x+2\right)\left(x-2\right)}\right)=\frac{2x-2}{\left(x+2\right)^2}\cdot\frac{\left(x+2\right)\left(x-2\right)}{-x}\)

\(=\frac{2\left(x-1\right)\left(x-2\right)}{-x\left(x+2\right)}\)

bạn ơi tới chừ bạn đã có lời giải chưa

Sửa đề :

a) \(A=\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{4}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-5}{x-\sqrt{x}-2}\right)\)

\(\Leftrightarrow A=\frac{x-\sqrt{x}+4\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-4-x+\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

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

b) \(A=4\)

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

\(\Leftrightarrow x+3\sqrt{x}+4=4\sqrt{x}+4\)

\(\Leftrightarrow x-\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy \(A=4\Leftrightarrow x\in\left\{0;1\right\}\)

\(\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2-\frac{x^2-10}{x+2}\right)\left(ĐK:x\ne\pm2\right)\)

\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{\left(x-2\right)\left(x+2\right)-\left(x^2-10\right)}{x+2}\)

\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{x^2-4-x^2+10}\)

\(=\frac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}=-\frac{1}{x-2}=\frac{1}{2-x}\)