\(\frac{x-1}{x}+\frac{x-2}{x+1}=2\)
ĐK: x ≠ 0; x ≠ -1
<=> \(\frac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\frac{x\left(x-2\right)}{x\left(x+1\right)}=\frac{2x\left(x+1\right)}{x\left(x+1\right)}\)
=> (x - 1)(x + 1) + x(x - 2) = 2x(x + 1)
<=> x2 - 1 + x2 - 2x = 2x2 + 2x
<=> 2x2 - 2x -1 -2x2 -2x = 0
<=> -4x - 1= 0
<=> x = \(\frac{-1}{4}\) (tmđk)