Ta có: \(\left(x^2-3x+2\right)\left(-x^2+5x-6\right)>=0\)
=>\(\left(x^2-3x+2\right)\left(x^2-5x+6\right)< =0\)
=>\(\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x-3\right)< =0\)
=>\(\left(x-1\right)\left(x-3\right)\cdot\left(x-2\right)^2< =0\)
TH1: \(\left(x-1\right)\left(x-3\right)\left(x-2\right)^2=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=2\end{matrix}\right.\)
TH2: \(\left(x-1\right)\left(x-3\right)\left(x-2\right)^2< 0\)
mà \(\left(x-2\right)^2>=0\forall x\)
nên \(\left(x-1\right)\left(x-3\right)< 0\)
=>1<x<3
Do đó: 1<=x<=3
=>a=1;b=3
a+b=4
