\(a,\)
\(B=\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}-\dfrac{4}{1-x^2}\) (Điều kiện xác định: \(x\ne\pm1\))
\(=\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{4}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2-2x+1-\left(x^2+2x+1\right)+4}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{-4x+4}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{-4\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=-\dfrac{4}{x+1}\)
\(b,\)
\(x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Với \(x=0\Rightarrow B=-\dfrac{4}{0+1}=-4\)
Với \(x=1\Rightarrow B=-\dfrac{4}{1+1}=-2\)
\(c,\)
\(B=-3\Rightarrow-\dfrac{4}{x+1}=-3\)
\(\Leftrightarrow-3\left(x+1\right)=-4\)
\(\Leftrightarrow-3x-3+4=0\)
\(\Leftrightarrow-3x+1=0\)
\(\Leftrightarrow-3x=-1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
\(d,\)
\(B< 0\Rightarrow-\dfrac{4}{x+1}< 0\)
\(\Leftrightarrow x+1>0\)
\(\Leftrightarrow x>-1\)
Kết hợp điều kiện \(x\ne\pm1\)
\(\Rightarrow-1< x< 1\)