Question

Tìm giá trị nhỏ nhất của biểu thức:\(2x^2+10x-1\)

Answer 1

\(2x^2+10x-1=\\ 2\left(x^2+5x-\dfrac{1}{2}\right)\\ =2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}-\dfrac{27}{4}\right)\\ =2\left(\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{4}\right)\\ =2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\)

vì \(2\left(x+\dfrac{5}{2}\right)^2\ge0\Rightarrow2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge-\dfrac{27}{2}\)vậy Min \(2x^2+10x-1\) \(=-\dfrac{27}{2}\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2=0\)

\(\Rightarrow x+\dfrac{5}{2}=0\Rightarrow x=-\dfrac{5}{2}\)