Question

\(\frac{x^2+x+1}{x^2+2x+1}+\frac{x^2+3x+1}{x^2+4x+1}=\frac{19}{12}\)

Answer 1

ĐKXĐ: ...

\(\Leftrightarrow\frac{x^2+x+1}{x^2+2x+1}-\frac{3}{4}+\frac{x^2+3x+1}{x^2+4x+1}-\frac{5}{6}=0\)

\(\Leftrightarrow\frac{x^2-2x+1}{4\left(x^2+2x+1\right)}+\frac{x^2-2x+1}{6\left(x^2+4x+1\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+1=0\\4\left(x^2+2x+1\right)=6\left(x^2+4x+1\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)