2/x-1+2x+3/x^2+x+1-(2x-1)(2x-1)/x^3-1 GIÚP VỚI

Đáp án:

$\dfrac{2}{x-1+2x}+\dfrac{3}{x^2+x+1}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{x^3-1}$ $=\dfrac{-10x^3+25x^2-19x+2}{3x^4-x^3-3x+1}$

Giải thích các bước giải:

$\dfrac{2}{x-1+2x}+\dfrac{3}{x^2+x+1}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{x^3-1}$ `;(x!=1/3;x!=1)`

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

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

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

$=\dfrac{-10x^3+25x^2-19x+2}{\left(x-1\right)\left(3x-1\right)\left(x^2+x+1\right)}$

$=\dfrac{-10x^3+25x^2-19x+2}{3x^4-x^3-3x+1}$