Giải thích các bước giải:
Ta có :
$C=\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}+\dfrac{1}{x+1}$
$\to C=\dfrac{\left(x^2+2\right)\left(x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}+\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}$
$\to C=\dfrac{\left(x^2+2\right)\left(x+1\right)+\left(x-1\right)\left(x+1\right)^2+\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}$
$\to C=\dfrac{3x^3+2x^2+x}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}$
b.Ta có :
$C-\dfrac{1}{3}=\dfrac{3x^3+2x^2+x}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}-\dfrac 13$
$\to C-\dfrac{1}{3}=\dfrac{-x^4+8x^3+6x^2+4x+1}{3\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}$
Không có nhận xét gì