a. \(\Rightarrow B=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{x+1-x+1}{x-1}\)
\(\Rightarrow B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(1-x\right)}.\dfrac{x-1}{2}\)
\(\Rightarrow B=\dfrac{x+\sqrt{x}+2\sqrt{x}+2-x+\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(1-x\right)}.\dfrac{-\left(1-x\right)}{2}\)
\(\Rightarrow B=\dfrac{-6\sqrt{x}}{2\left(\sqrt{x}-1\right)}=\dfrac{-3\sqrt{x}}{\sqrt{x}-1}\) (1)
b. Ta có: \(\left|2\sqrt{x}-1\right|=3\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}-1=3\\2\sqrt{x}-1=-3\end{matrix}\right.\Leftrightarrow x=4\)
Thay x=4 vào (1), ta có:\(\dfrac{-3.\sqrt{4}}{\sqrt{4}-1}=-6\)
Vậy gtbt B với \(\left|2\sqrt{x}-1\right|=3\) là -6
