Question

chứng tỏ rằng 9x^2-6x+3>0 với mọi x thuộc  R

Answer 1

      \(9x^2-6x+3\)

\(=\left(9x^2-6x+1\right)+2\)

\(=\left(3x-1\right)^2+2\)

Vì    \(\left(3x-1\right)^2\ge0\)

\(\Rightarrow\)\(\left(3x-1\right)^2+2>0\)

hay    \(9x^2-6x+1>0\)

Answer 2

Ta có :

\(9x^2-6x+3\)

\(=\left(9x^2-6x+1\right)+2\)

\(=\left(3x-1\right)^2+2\)

Mà \(\left(3x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow\left(3x-1\right)^2+2\ge2>0\forall x\in R\)

Vậy \(9x^2-6x+3>0\forall x\in R\)