`a)`\(x^2-3x+11=x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2+11=\left(x-\dfrac{3}{2}\right)^2+\dfrac{35}{4}\ge\dfrac{35}{4}>0;\forall x\)
`b)`\(x^2-\dfrac{2}{3}x+1=x^2-2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^2+1=\left(x-\dfrac{1}{3}\right)^2+\dfrac{8}{9}\ge\dfrac{8}{9}>0;\forall x\)
`c)`\(2x^2-5x+18=2\left(x^2-\dfrac{5}{2}x+9\right)=2\left(x^2-2.x.\dfrac{5}{4}+\left(\dfrac{5}{4}\right)^2-\left(\dfrac{5}{4}\right)^2+9\right)=2\left[\left(x-\dfrac{5}{4}\right)^2\right]+\dfrac{119}{8}\ge\dfrac{119}{8}>0;\forall x\)
`d)`\(\dfrac{2}{3}x^2-x+5=\dfrac{2}{3}\left(x^2-\dfrac{3}{2}x+\dfrac{15}{2}\right)=\dfrac{2}{3}\left(x^2-2.x.\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2-\left(\dfrac{3}{4}\right)^2+\dfrac{15}{2}\right)=\dfrac{2}{3}\left(x-\dfrac{3}{4}\right)^2+\dfrac{37}{8}\ge\dfrac{37}{8}>0;\forall x\)
a: \(=x^2-3x+\dfrac{9}{4}+\dfrac{33}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{33}{4}>0\)
b: \(=x^2-\dfrac{2}{3}x+\dfrac{1}{9}+\dfrac{8}{9}=\left(x-\dfrac{1}{3}\right)^2+\dfrac{8}{9}>0\)
c: \(=2\left(x^2-\dfrac{5}{2}x+9\right)\)
\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{119}{16}\right)\)
\(=2\left(x-\dfrac{5}{4}\right)^2+\dfrac{119}{8}>0\)
