\(2,A=x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}-\dfrac{9}{4}\\ =\left(x-\dfrac{5}{4}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\forall x\)
Dấu = xảy ra khi
\(x-\dfrac{5}{4}=0\\ x=\dfrac{5}{4}\)
Vậy \(Min_A=-\dfrac{9}{4}khix=\dfrac{5}{4}\)
3,
\(M=-\left(x^2+3x-4\right)\\ =-\left(x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{25}{4}\right)\\ =-\left(x+\dfrac{2}{3}\right)^2-\dfrac{25}{4}\le-\dfrac{25}{4}\)
Dấu = xảy ra khi
\(x+\dfrac{2}{3}=0\\ x=-\dfrac{2}{3}\)
Vậy \(Min_M=-\dfrac{25}{4}khix=\dfrac{-2}{3}\)
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