đề bài là như vậy hả ???
\(\left(\dfrac{1-4}{\sqrt{x}-1}+\dfrac{1}{x+1}\right).\dfrac{x-2\sqrt{x}}{x-1}\)
a:\(A=\left(1-\dfrac{4}{\sqrt{x}+1}+\dfrac{1}{x-1}\right):\dfrac{x-2\sqrt{x}}{x-1}\)
\(=\dfrac{x-1-4\sqrt{x}+4+1}{x-1}\cdot\dfrac{x-1}{x-2\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b: Để A=1/2 thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{2}\)
=>2 căn x-4=căn x
=>x=16