Question

Tìm giá trị nhỏ nhất của biểu thức:

A=2x² + 14x + 15

Answer 1

A=\(2x^2+14x+15=2x^2+2.\sqrt{2}x.\dfrac{7}{\sqrt{2}}+\dfrac{49}{2}-\dfrac{19}{2}\)

=\(\left(\sqrt{2}x+\dfrac{7}{\sqrt{2}}\right)^2-\dfrac{19}{2}\)

Do\(\left(\sqrt{2}x+\dfrac{7}{\sqrt{2}}\right)^2\)>=0

<=>\(\left(\sqrt{2}x+\dfrac{7}{\sqrt{2}}\right)^2\)\(-\dfrac{19}{2}\)>=\(\dfrac{-19}{2}\)

Dấu"=" xảy ra khi\(\left(\sqrt{2}x+\dfrac{7}{\sqrt{2}}\right)^2\)=0

<=>\(\left(\sqrt{2}x+\dfrac{7}{\sqrt{2}}\right)\)=0

<=>\(\sqrt{2}x=\dfrac{-7}{\sqrt{2}}\)

<=>\(x=\dfrac{-7}{2}\)

Vậy MinA=\(\dfrac{-19}{2}\)khi \(x=\dfrac{-7}{2}\)