Pt tương đương:
\(\dfrac{x^2+2x+1-x}{x^2+2x+1}+\dfrac{x^2+4x+1-x}{x^2+4x+1}=\dfrac{19}{12}\Leftrightarrow1-\dfrac{x}{x^2+2x+1}+1-\dfrac{x}{x^2+4x+1}=\dfrac{19}{12}\)
\(\Leftrightarrow-\dfrac{x}{x^2+2x+1}-\dfrac{x}{x^2+4x+1}+\dfrac{5}{12}=0\)
\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{x}{x^2+2x+1}\right)+\left(\dfrac{1}{6}-\dfrac{x}{x^2+4x+1}\right)=0\)
\(\Leftrightarrow\dfrac{x^2-2x+1}{4\left(x^2+2x+1\right)}+\dfrac{x^2-2x+1}{6\left(x^2+4x+1\right)}=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(\dfrac{1}{4\left(x^2+2x+1\right)}+\dfrac{1}{6\left(x^2+4x+1\right)}\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\dfrac{\left(10x^2+32x+10\right)}{24\left(x+1\right)^2\left(x^2+4x+1\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\10x^2+32x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-8+\sqrt{39}}{5}\\x=\dfrac{-8-\sqrt{39}}{5}\end{matrix}\right.\)
