Question

Tìm giá trị nhỏ nhất của biểu thức:

D= 2x^2 - 6x

Answer 1

\(D=2x^2-6x\)

\(=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\)

\(=2\left[x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{9}{2}\)

\(=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)

Ta có :

\(2\left(x-\dfrac{3}{2}\right)^2\ge0\Rightarrow2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

hay D ≥ \(-\dfrac{9}{2}\)

Dấu = xảy ra \(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)

Vậy \(Min_D=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)