Giải các phương trình: x+1/x^2+x+1 -x-1/x^2-x+1=3/x(x^4+x+1)

Đáp án: $x=\dfrac32$

Giải thích các bước giải:

Ta có:

$\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^4+x^2+1)}$

$\to \dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^4+2x^2+1-x^2)}$

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

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

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

$\to x\left(x+1\right)\left(x^2-x+1\right)-x\left(x-1\right)\left(x^2+x+1\right)=3$

$\to x(x^3+1)-x(x^3-1)=3$

$\to (x^4+x)-(x^4-x)=3$

$\to 2x=3$

$\to x=\dfrac32$