TH1: x>=2
BPT sẽ là x-2<=2x^2-9x+9
=>2x^2-9x+9>=x-2
=>2x^2-10x+11>=0
=>\(\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=\dfrac{5+\sqrt{3}}{2}\end{matrix}\right.\Leftrightarrow x>=\dfrac{5+\sqrt{3}}{2}\)
TH2: x<2
BPT sẽ là 2x^2-9x+9>=-x+2
=>2x^2-8x+7>=0
\(\Leftrightarrow x< =\dfrac{4-\sqrt{2}}{2}\)