Question

 

\(\text{Tìm gtnn của: }2x^2-3x+1\)

Answer 1

\(2x^2-3x+1\)

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

\(=2\left[\left(x^2-2.x.\frac{3}{4}+\frac{9}{16}\right)-\frac{9}{16}+\frac{1}{2}\right]\)

\(=2\left[\left(x-\frac{3}{4}\right)^2-\frac{1}{16}\right]\)

\(=2\left(x-\frac{3}{4}\right)^2-\frac{1}{8}\ge\frac{1}{8}\)

Vậy \(Min\left(2x^2-3x+1\right)=\frac{1}{8}\)

Answer 2

Ta có \(2x^2\ge0\)giả sử x=0

           \(-3x\ge0\)giả sử x cũng= 0

          1>0

\(\Rightarrow2x^2-3x+1\ge1\)

vậy gtnn của:\(2x^2-3x+1\)là 1