\(\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}=\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}\) \(\left(dkxd:x\ne-2\right)\)
\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}-\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(3x+1\right)-x\right]=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(6x+2-x\right)=0\)
Bỏ vế đằng trước \(x^2+x+1=0\) do vô nghiệm
\(\Leftrightarrow6x+2-x=0\)
\(\Leftrightarrow5x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{5}\left(tmdk\right)\)
Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
\(\dfrac{\left(x^2+x+1\right).\left(3x+1\right)}{x+2}=\dfrac{\left(x^2+x+1\right).x}{2\left(x+2\right)}\)
hay \(\left(x^2+x+1\right).\dfrac{3x+1}{x+2}=\left(x^2+x+1\right).\dfrac{x}{2\left(x+2\right)}\)