Ta có: \(\frac{9x^2-12x+4}{x^2-2x+2}\)
= \(\frac{9\left(x^2-2x+2\right)+6x-14}{x^2-2x+2}\)
= \(9+\frac{6x-14}{x^2-2x+2}\)
= \(9+\frac{x^2-2x+2-\left(x^2-8x+16\right)}{\left(x^2-2x+1\right)+1}\)
= \(9+1-\frac{\left(x-4\right)^2}{\left(x-1\right)^2+1}\)
= \(10-\frac{\left(x-4\right)^2}{\left(x-1\right)^2+1}\le10\forall x\)
Dấu "=" xảy ra <=> x - 4 = 0 <=> x = 4
Vậy Max của \(\frac{9x^2-12x+4}{x^2-2x+2}\)= 10 khi x = 4