a)Đa thức B có nghĩa\(\Leftrightarrow x+1\ne0\)và\(x-1\ne0\)và\(x\ne0\Leftrightarrow x\ne-1\)và\(x\ne1\)và\(x\ne0\)
b)Ta có:\(B=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(\frac{x^2+1}{x+1}-\frac{x+1}{x+1}\right)\left(\frac{4.x}{\left(x-1\right).x}-\frac{2.\left(x-1\right)}{x.\left(x-1\right)}\right)\)
\(=\frac{x^2+1-x-1}{x+1}\left(\frac{4x}{x\left(x-1\right)}-\frac{2x-2}{x\left(x-1\right)}\right)=\frac{x^2-x}{x+1}.\frac{4x-2x+2}{x\left(x-1\right)}=\frac{x\left(x-1\right)}{x+1}.\frac{2x+2}{x\left(x-1\right)}\)
\(=\frac{2x+2}{x+1}=\frac{2\left(x+1\right)}{x+1}=2\)
a) B có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+1\ne0\\x-1\ne0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne1\\x\ne0\end{cases}}\)
b) \(B=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(=\frac{\left(x^2+1\right)-\left(x+1\right)}{x+1}.\frac{4x-\left(2x-2\right)}{x\left(x-1\right)}\)
\(=\frac{x^2+1-x-1}{x+1}.\frac{4x-2x+2}{x\left(x-1\right)}\)
\(=\frac{x^2-x}{x+1}.\frac{2x+2}{x\left(x-1\right)}=\frac{x\left(x-1\right)}{x+1}.\frac{2\left(x+1\right)}{x\left(x-1\right)}=2\)
