Question

giải phương trình sau:

\(\frac{1}{x-1}+\frac{3x^2}{1-x^3}=\frac{2x}{x^2+x+1}\)

Answer 1

ĐKXĐ: \(x\ne1\)

\(\frac{1}{x-1}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x}{x^2+x+1}=0\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow x^2+x+1-3x^2-2x\left(x-1\right)=0\)

\(\Leftrightarrow-4x^2+3x+1=0\)

\(\Leftrightarrow\left(1-x\right)\left(4x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=-\frac{1}{4}\end{matrix}\right.\)