\(ĐK:x\ge1\\ PT\Leftrightarrow x^2-3x-1=x-1\\ \Leftrightarrow x^2-4x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\)
\(B=\left[\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\sqrt{x}+4\right)}\cdot\dfrac{x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\cdot\dfrac{\sqrt{x}+2}{4}\\ B=\left[\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x+2\sqrt{x}+4}{\left(\sqrt{x}+2\right)^2}\right]\cdot\dfrac{\sqrt{x}+2}{4}\\ B=\dfrac{x+2\sqrt{x}-x-2\sqrt{x}-4}{\left(\sqrt{x}+2\right)^2}\cdot\dfrac{\sqrt{x}+2}{4}\\ B=\dfrac{-4}{4\left(\sqrt{x}+2\right)}=\dfrac{-1}{\sqrt{x}+1}\)
