\(\dfrac{2x^2-1}{x^3+1}+\dfrac{1}{x+1}=2x\left(1-\dfrac{x^2-x}{x^2-x+1}\right)\)
\(\Leftrightarrow\dfrac{2x^2-1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x+1}=2x\left(1-\dfrac{x^2-x}{x^2-x+1}\right)\)
ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x^2-x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\end{matrix}\right.\)
Ta có : \(\dfrac{2x^2-1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x+1}=2x\left(1-\dfrac{x^2-x}{x^2-x+1}\right)\)
\(\Leftrightarrow\dfrac{2x^2-1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=2x\left(1-\dfrac{x^2-x}{x^2-x+1}\right)\)
\(\Leftrightarrow\dfrac{2x^2-1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=2x-\dfrac{2x\left(x^2-x\right)}{x^2-x+1}\)
\(\Leftrightarrow\dfrac{2x^2-1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x\left(x^2-x+1\right)}{x^2-x+1}-\dfrac{2x\left(x^2-x\right)}{x^2-x+1}\)\(\Leftrightarrow\dfrac{2x^2-1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x^3-2x^2+2x-2x^3+2x^2}{x^2-x+1}\)\(\Leftrightarrow\dfrac{2x^2-1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x}{x^2-x+1}\)
\(\Leftrightarrow\dfrac{2x^2-1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
`=> 2x^2 -1 + x^2 -x+1 - 2x^2 -2x=0`
`<=> x^2 -3x=0`
`<=> x(x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{0;-3\right\}\)
