Question

\(\frac{1}{x-1}\) + \(\frac{2x^2-5}{x^3-1}\) = \(\frac{4}{x^2+x+1}\)

giải phương trình

Answer 1

\(\Leftrightarrow\frac{x^2+x+1}{x^3-1}+\frac{2x^2-5}{x^3-1}=\frac{4\left(x-1\right)}{x^3-1}\)

\(\Leftrightarrow x^2+x+1+2x^2-5=4x-4\)

\(\Leftrightarrow3x^2-3x=0\Leftrightarrow3x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Answer 2

\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)

\(\frac{x^2+x+1}{x^3-1}+\frac{2x^2-5}{x^3-1}-\frac{4\left(x-1\right)}{x^3-1}=0\)

\(\frac{x^2+x+1+2x^2-5-4x+4}{x^3-1}=0\)

\(3x^2-3x=0\)

\(3x\left(x-1\right)=0\)

\(\left[{}\begin{matrix}3x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)