\(\dfrac{1}{x^2-3x+3}+\dfrac{2}{x^2-3x+4}=\dfrac{6}{x^2-3x+5}\)
Đặt \(x^2-3x+3=t\) , ta có :
\(\dfrac{1}{t}+\dfrac{2}{t+1}=\dfrac{6}{t+2}\)
\(\Leftrightarrow\dfrac{1}{t}+\dfrac{2}{t+1}-\dfrac{6}{t+2}=0\)
\(\Leftrightarrow\dfrac{\left(t+1\right)\left(t+2\right)+2t\left(t+2\right)-6t\left(t+1\right)}{t\left(t+1\right)\left(t+2\right)}=0\)
\(\Leftrightarrow\dfrac{t^2+3t+2+2t^2+4t-6t^2-6t}{t\left(t+1\right)\left(t+2\right)}=0\)
\(\Leftrightarrow\dfrac{-3t^2+t+2}{t\left(t+1\right)\left(t+2\right)}=0\)
\(\Leftrightarrow-3t^2+t+2=0\)
\(\Leftrightarrow-3t^2+3t-2t+2=0\)
\(\Leftrightarrow-3t\left(t-1\right)-2\left(t-1\right)=0\)
\(\Leftrightarrow-\left(t-1\right)\left(3t+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-1=0\\3t+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{2}{3}\end{matrix}\right.\)
Với t = 1
\(\Rightarrow x^2-3x+3=1\)
\(\Leftrightarrow x^2-3x+3-1=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-2x-x+2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Với \(t=-\dfrac{2}{3}\)
\(\Rightarrow x^2-3x+3=-\dfrac{2}{3}\)
\(\Leftrightarrow x^2-3x+3+\dfrac{2}{3}=0\)
\(\Leftrightarrow x^2-3x+\dfrac{11}{3}=0\)
Ta có :
\(x^2-3x+\dfrac{11}{3}\)
\(=\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{4}+\dfrac{11}{3}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{17}{12}\)
Ta có : \(\left(x-\dfrac{3}{2}\right)^2\ge0\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{17}{12}\ge\dfrac{17}{2}>0\) với mọi x
\(\Rightarrow x^2-3x+\dfrac{11}{3}=0\) ( vô lý )
Vậy tập nghiệm của phương trình \(S=\left\{1;2\right\}\)