\(9x^2-3x+2=9\left(x^2-\dfrac{1}{3}x+\dfrac{2}{9}\right)\)
\(=9\left[x^2-2.x.\dfrac{1}{6}+\left(\dfrac{1}{6}\right)^2-\dfrac{1}{36}+\dfrac{2}{9}\right]\)
\(=9\left[\left(x-\dfrac{1}{6}\right)^2+\dfrac{7}{36}\right]\)
\(=9\left(x-\dfrac{1}{6}\right)^2+\dfrac{7}{4}\)
Do \(9\left(x-\dfrac{1}{6}\right)^2\ge0\) với \(\forall x\)
\(\Rightarrow9\left(x-\dfrac{1}{6}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\) hay \(9x^2-3x+2>0\) với mọi x
=> Điều phải chứng minh
9x2-3x+2
=[(3x)2-2.3x.\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)]+\(\dfrac{7}{4}\)
=(3x-\(\dfrac{1}{2}\))2+\(\dfrac{7}{4}\)
Vì (3x-\(\dfrac{1}{2}\))2\(\ge\)0
=> (3x-\(\dfrac{1}{2}\))2+\(\dfrac{7}{4}\)>0∀x
Đúng thì tick nha,
