\(\text{ĐKXĐ}:x\ne1\)
\(\dfrac{1}{x-1}+\dfrac{2}{x^2+x+1}=\dfrac{3x^2}{x^3-1}\)
\(\Leftrightarrow\dfrac{1}{x-1}+\dfrac{2}{x^2+x+1}-\dfrac{3x^2}{x^3-1}=0\)
\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)+2\left(x-1\right)-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Rightarrow\left(x^2+x+1\right)+2\left(x-1\right)-3x^2=0\)
\(\Leftrightarrow x^2+x+1+2x-2-3x^2=0\)
\(\Leftrightarrow-2x^2+3x-1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{1}{2};1\right\}\)
đk : x ≠ 1
\(\dfrac{1\left(x^2+x+1\right)+2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\\ x^2+x+1+2x-2-3x^2=0\\ -2x^2+3x-1=0\)
\(-2x^2+2x+x-1=0\\ -2x\left(x-1\right)+\left(x-1\right)=0\\ \left(-2x+1\right)\left(x-1\right)=0\)
=> x= 1 ; \(x=\dfrac{1}{2}\)
