1. Ta có: \(f\left(x\right)=9x^2-12x+1=\left(3x\right)^2-2.3x.2+2^2-3\)
\(=\left(3x-2\right)^2-3\)
Vì \(\left(3x-2\right)^2\ge0\) với mọi x \(\Rightarrow\left(3x-2\right)^2-3\ge-3\) hay \(f\left(x\right)\ge-3\)
Dấu ''='' xảy ra \(\Leftrightarrow\left(3x-2\right)^2=0\Rightarrow3x-2=0\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
Vậy min f(x) =-3 khi \(x=\dfrac{2}{3}\)
2. Ta có: \(f\left(x\right)=2x^2-7x+5=2.\left(x^2-3,5x\right)+5=2.\left(x^2-2.x.1,75+1,75^2\right)-2.1,75^2+5\)
\(=2.\left(x-1,75\right)^2-1,125\)
Vì \(2.\left(x-1,75\right)^2\ge0\Rightarrow2.\left(x-1,75\right)^2-1,125\ge-1,125\Rightarrow f\left(x\right)\ge-1,125\)
Dấu ''='' xảy ra \(\Leftrightarrow2.\left(x-1,75\right)^2=0\Rightarrow x-1,75=0\Rightarrow x=1,75\)
Vậy min f(x)=-1,125 khi x=1,75
3.\(3x^2-10x=3.\left(x^2-\dfrac{10}{3}x\right)=3.\left(x^2-2.x.\dfrac{5}{3}\right)\)
\(=3.\left[x^2-2.x.\dfrac{5}{3}+\left(\dfrac{5}{3}\right)^2\right]-3.\left(\dfrac{5}{3}\right)^2\)
\(=3.\left(x-\dfrac{5}{3}\right)^2-\dfrac{25}{3}\)
Vì \(3.\left(x-\dfrac{5}{3}\right)^2\ge0\Rightarrow3.\left(x-\dfrac{5}{3}\right)^2-\dfrac{25}{3}\ge-\dfrac{25}{3}\Rightarrow f\left(x\right)\ge-\dfrac{25}{3}\)
Dấu ''='' xảy ra \(\Leftrightarrow3.\left(x-\dfrac{5}{3}\right)^2=0\Rightarrow x-\dfrac{5}{3}=0\Rightarrow x=\dfrac{5}{3}\)
Vậy min f(x)=\(-\dfrac{25}{3}\) khi \(x=\dfrac{5}{3}\)
