a) ĐKXĐ \(\left\{{}\begin{matrix}x\ge0\\x\ne1;4\end{matrix}\right.\)
\(M=\left(1-\frac{4\sqrt{x}}{x-1}+\frac{1}{\sqrt{x}-1}\right):\frac{x-2\sqrt{x}}{x-1}\\ =\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\left(\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\frac{x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
b)
\(M=\frac{\sqrt{x}-3}{\sqrt{x}-2}=\frac{1}{2}\\ \Leftrightarrow\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{1}{2}=0\\ \Leftrightarrow\frac{2\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}-2}{2\left(\sqrt{x}-2\right)}=0\\ \Leftrightarrow\frac{\sqrt{x}-4}{2\left(\sqrt{x}-2\right)}=0\\ \Leftrightarrow\sqrt{x}-4=0\Leftrightarrow x=16\)
(Bạn kiểm tra phần rút gọn với đáp án cho chắc nha)
