\(a.B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\dfrac{\sqrt{x}+1-\sqrt{x}-1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{1-x}\left(x\ge0;x\ne1\right)\)
\(b.Với:x=3\left(TMĐK\right)\) , ta có :
\(B=\dfrac{\sqrt{3}}{1-3}=-\dfrac{\sqrt{3}}{2}\)
\(c.\left|B\right|=\dfrac{1}{2}\Leftrightarrow\left|\dfrac{\sqrt{x}}{1-x}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{x}}{1-x}=\dfrac{1}{2}\\\dfrac{\sqrt{x}}{1-x}=-\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2\sqrt{x}+x-1}{2\left(1-x\right)}=0\\\dfrac{2\sqrt{x}-x+1}{2\left(1-x\right)}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2\sqrt{x}+1-2=0\\x-2\sqrt{x}+1-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left(\sqrt{x}+1\right)^2-2=0\\\left(\sqrt{x}-1\right)^2-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\sqrt{x}+1-\sqrt{2}\right)\left(\sqrt{x}+1+\sqrt{2}\right)=0\\\left(\sqrt{x}-1-\sqrt{2}\right)\left(\sqrt{x}-1+\sqrt{2}\right)=0\end{matrix}\right.\)
Tới đây dễ rồi , bn tự làm nốt nhé.
