\(\dfrac{1}{x^2-3x+3}+\dfrac{2}{x^2-3x+4}=\dfrac{6}{x^2-3x+5}\)
\(\Leftrightarrow1-\dfrac{1}{x^2-3x+3}+1-\dfrac{2}{x^2-3x+4}=2-\dfrac{6}{x^2-3x+5}\)
\(\Leftrightarrow\dfrac{x^2-3x+2}{x^2-3x+3}+\dfrac{x^2-3x+2}{x^2-3x+4}=\dfrac{2x^2-6x+4}{x^2-3x+5}\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(\dfrac{1}{x^2-3x+3}+\dfrac{1}{x^2-3x+4}-\dfrac{2}{x^2-3x+5}\right)=0\left(1\right)\)
Nhận xét:
\(\left\{{}\begin{matrix}x^2-3x+3< x^2-3x+5\\x^2-3x+4< x^2-3x+5\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x^2-3x+3}>\dfrac{1}{x^2-3x+5}\\\dfrac{1}{x^2-3x+4}>\dfrac{1}{x^2-3x+5}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{x^2-3x+3}+\dfrac{1}{x^2-3x+4}>\dfrac{2}{x^2-3x+5}\)
\(\Rightarrow\dfrac{1}{x^2-3x+3}+\dfrac{1}{x^2-3x+4}-\dfrac{2}{x^2-3x+5}>0\)
Suy ra \(\left(1\right)\Rightarrow\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình \(S=\left\{1;2\right\}\)