Cho B=1/(x^2-x+1)+1-(x^2+2)/(x^3+1). Sau khi thu gọn hoàn toàn

* Hướng dẫn giải

Ta có B = $\frac{1}{{\mathrm{x}}^2-\mathrm{x}+1}+1-\frac{{\mathrm{x}}^2+2}{{\mathrm{x}}^3+1}$

=$\frac{1}{{\mathrm{x}}^2-\mathrm{x}+1}-\frac{{\mathrm{x}}^2+2}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}+1$

=$\frac{\mathrm{x}+1}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}-\frac{{\mathrm{x}}^2+2}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}+\frac{{\mathrm{x}}^3+1}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}$

=$\frac{\mathrm{x}+1-{\mathrm{x}}^2-2+{\mathrm{x}}^3+1}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}=\frac{{\mathrm{x}}^3-{\mathrm{x}}^2+\mathrm{x}}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}$

=$\frac{\mathrm{x}({\mathrm{x}}^2-\mathrm{x}+1)}{(\mathrm{x}+1\left)\right({\mathrm{x}}^2-\mathrm{x}+1)}=\frac{\mathrm{x}}{\mathrm{x}+1}$

Đáp án cần chọn là: A