ta có Q= x2-5x= x2-2x\(\frac{5}{2}\)+ \(\frac{25}{4}\)- \(\frac{25}{4}\)= (x-\(\frac{5}{2}\))2-\(\frac{25}{4}\)
vì (x-\(\frac{5}{2}\)) 2>=0
=> Q >= \(\frac{-25}{4}\)
dấu '=' sảy ra khi (x-\(\frac{5}{2}\))2=0
=> x-\(\frac{5}{2}\)=0
=>x=\(\frac{5}{2}\)
vậy Q(min)=\(\frac{-25}{4}\) khi x= \(\frac{5}{2}\)
Ta có :
\(Q=x^2+5x\)
\(\Rightarrow Q=x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}\)
\(\Rightarrow Q=\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\)
Vì \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\ge-\frac{25}{4}\)
Dấu " = " xảy ra khi x = - 5 / 4
Vậy ......
\(Q=x^2-5x=x^2-2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}\)
\(\Leftrightarrow Q=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\)
Vì \(\left(x-\frac{5}{2}\right)^2\ge O\Rightarrow Q\ge-\frac{25}{4}\)
Dấu ''='' xảy ra khi \(\left(x-\frac{5}{2}\right)^2=0\Leftrightarrow x=\frac{5}{2}\)
Vậy GTNN của Q = \(\frac{-25}{4}\Leftrightarrow x=\frac{5}{2}\)
