\(Q=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(\frac{x^2+1-\left(x+1\right)}{x+1}\right)\left(\frac{4x-2\left(x-1\right)}{x\left(x-1\right)}\right)\)
\(=\left(\frac{x^2+1-x-1}{x+1}\right)\left(\frac{4x-2x+2}{x\left(x-1\right)}\right)=\left(\frac{x^2-x}{x+1}\right)\left(\frac{2\left(x+1\right)}{x\left(x-1\right)}\right)=\frac{2x\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}=2\)
Vậy Q = 2
Hình như đề là rút gọn thì phải.
Giải
\(Q=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(=\left(\frac{x^2}{x}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(x-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(=\frac{4\left(x-1\right)}{x-1}-\frac{2\left(x-1\right)}{x}=4-\frac{2x-2}{x}\)
