Đặt t=x2-2x+3(t\(\ge\)2)
PTTT: \(\dfrac{1}{t-1}+\dfrac{1}{t}=\dfrac{9}{2\left(t+1\right)}\)
<=>2t2+2t+2t2-2=9t2-9
<=>5t2-2t-7=0
<=>(t+1)(5t-7)=0
Do t\(\ge\)2
=>t+1>0 5t-7>0
Vậy pt vô nghiệm
\(\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2-2x+3}=\dfrac{9}{2\left(x^2-2x+4\right)}\)
Đặt \(t=x^2-2x+2=\left(x-1\right)^2+1\ge1\)
Thì ta có:
\(PT\Leftrightarrow\dfrac{1}{t}+\dfrac{1}{t+1}=\dfrac{9}{2\left(t+2\right)}\)
\(\Leftrightarrow5t^2-t-4=0\)
\(\Leftrightarrow\left(5t^2-5t\right)+\left(4t-4\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left(5t+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5t+4=0\\t-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{4}{5}\left(l\right)\\t=1\end{matrix}\right.\)
\(\Rightarrow x^2-2x+2=1\)
\(\Leftrightarrow x=1\)
Vậy PT có 1 nghiệm là x = 1
