Question

giải phương trình: \(\dfrac{x-1}{x^2-x+1}-\dfrac{x+1}{x^2+x+1}=\dfrac{10}{x\left(x^4+x^2+1\right)}\)

Answer 1

Ta có:

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

<=> \(\dfrac{\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{10}{x\left(x^4+x^2+1\right)}\)

<=> \(\dfrac{x^3-1-x^3-1}{x^4+x^2+1}=\dfrac{10}{x\left(x^4+x^2+1\right)}\)

<=> \(\dfrac{-2}{x^4+x^2+1}=\dfrac{10}{x\left(x^4+x^2+1\right)}\)

<=> \(\dfrac{-2x}{x\left(x^4+x^2+1\right)}=\dfrac{10}{x\left(x^4+x^2+1\right)}\)

=> ^_ 2x = 10

<=> x =-5

Vậy x = -5