Question

\(\dfrac{ }{ }\)\(\dfrac{2}{x^2+2}+\dfrac{3x^2-6x}{x^2-2x+4}+\dfrac{10x^2+28-8}{x^4+8x}\)

Answer 1

Sửa đề: \(\dfrac{2}{x+2}+\dfrac{3x^2-6x}{x^2-2x+4}+\dfrac{10x^2+28x-8}{x^4+8x}\)

\(=\dfrac{2}{x+2}+\dfrac{3x\left(x-2\right)}{x^2+2x+4}+\dfrac{10x^2+28x-8}{x\left(x+2\right)\left(x^2-2x+4\right)}\)

\(=\dfrac{2x\left(x^2+2x+4\right)+3x^2\left(x^2-4\right)+10x^2+28x-8}{x\left(x+2\right)\left(x^2-2x+4\right)}\)

\(=\dfrac{2x^3+4x^2+8x+3x^4-12x^2+10x^2+28x-8}{x\left(x+2\right)\left(x^2-2x+4\right)}\)

\(=\dfrac{3x^4+2x^3+2x^2+36x-8}{x\left(x+2\right)\left(x^2-2x+4\right)}\)