Tính: x-1/x^2 / 1+ 1/x + 1/x^2

* Hướng dẫn giải

$\begin{array}{l}\frac{\mathrm{x}-\frac{1}{{\mathrm{x}}^2}}{1+\frac{1}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}}=\left(\mathrm{x}-\frac{1}{{\mathrm{x}}^2}\right):\left(1+\frac{1}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}\right)=\frac{{\mathrm{x}}^3-1}{{\mathrm{x}}^2}:\frac{{\mathrm{x}}^2+\mathrm{x}+1}{{\mathrm{x}}^2}\\ =\frac{\left(\mathrm{x}-1\right)\left({\mathrm{x}}^2+\mathrm{x}+1\right)}{{\mathrm{x}}^2}.\frac{{\mathrm{x}}^2}{{\mathrm{x}}^2+\mathrm{x}+1}=\mathrm{x}-1\end{array}$