\(\frac{6}{x^3+1}-\frac{1-x}{x^2-x+1}=\frac{5}{x+1}\)
\(\Leftrightarrow\frac{6}{\left(x+1\right)\left(x^2-x.1+1^2\right)}-\frac{1-x}{x^2-x+1}=\frac{5}{x+1}\)
\(\Leftrightarrow\frac{6}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{1-x}{x^2-x+1}=\frac{5}{x+1}\)
\(\Leftrightarrow6-\left(1-x\right)\left(x+1\right)=5\left(x^2-x+1\right)\)
\(\Leftrightarrow x^2+5=5x^2-5x+5\)
\(\Leftrightarrow x^2=5x^2-5x\)
\(\Leftrightarrow x^2-5x^2+5x=0\)
\(\Leftrightarrow-4x^2+5x=0\)
\(\Leftrightarrow-x\left(4x-5\right)=0\)
\(\Leftrightarrow x\left(4x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{4}\end{matrix}\right.\)
Nghiệm phương trình là \(\left\{0;\frac{5}{4}\right\}\)
