Question

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

Answer 1

ĐKXĐ: \(x\ne1\)

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

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

=>\(2x^2+2x+2+2x^2-2x+3x-3=4x^2-1\)

=>3x-1=-1

=>3x=0

=>x=0(nhận)