\(\left(x^2-3x+1\right)^2+3x^2-9x+5>0\)
=>\(\left(x^2-3x+1\right)^2+3x^2-9x+3+2>0\)
=>\(\left(x^2-3x+1\right)^2+3\left(x^2-3x+1\right)+2>0\)
=>\(\left(x^2-3x+1+1\right)\left(x^2-3x+1+2\right)>0\)
=>\(\left(x^2-3x+2\right)\left(x^2-3x+3\right)>0\)
mà \(x^2-3x+3=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{3}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
nên \(x^2-3x+2>0\)
=>(x-2)(x-1)>0
TH1: \(\left\{{}\begin{matrix}x-2>0\\x-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>2\\x>1\end{matrix}\right.\)
=>x>2
TH2: \(\left\{{}\begin{matrix}x-2< 0\\x-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 2\\x< 1\end{matrix}\right.\)
=>x<1
