Question

Tìm GTNN của biểu thức Q = x^{2  -}  5x

Answer 1

Q=x².5x

=x²-5x+\(\frac{25}{4}-\frac{25}{4}\)

=\(\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\)

\(\Rightarrow\left(x-\frac{5}{2}\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{5}{2}\right)^2\ge-\frac{25}{4}\)

Vay GTNN la : \(-\frac{25}{4}\)

Dấu "=" xảy ra khi : \(x-\frac{5}{2}=0\Rightarrow x=\frac{5}{2}\)

Answer 2

MinQ=\(\frac{-\Delta}{4a}=\frac{-b^2+4ac}{4a}=\frac{-5^2+4.1.0}{4}=-\frac{25}{4}\)