Đặt \(A=-5x^2-4x+1\)
\(=-\left(5x^2+4x-1\right)\)
\(=-\left(5x^2+2\cdot x\cdot2+2^2-4-1\right)\)
\(=-\left[\left(x+2\right)^2-5\right]\)
\(=-\left(x+2\right)^2+5\le5\)
Dấu "=" xảy ra khi \(x=-2\)
Vậy \(MAX_A=5\) tại \(x=-2\)
a, Đặt \(A=-5x^2-4x+1\)
\(=-5\left(x^2+\dfrac{4}{5}x-\dfrac{1}{5}\right)\)
\(=-5\left(x^2+2.x.\dfrac{2}{5}+\dfrac{4}{25}-\dfrac{9}{25}\right)\)
\(=-5\left[\left(x+\dfrac{2}{5}\right)^2-\dfrac{9}{25}\right]\)
\(=-5\left(x+\dfrac{2}{5}\right)^2+\dfrac{9}{5}\le\dfrac{9}{5}\)
Dấu " = " khi \(-5\left(x+\dfrac{2}{5}\right)^2=0\Leftrightarrow x=\dfrac{-2}{5}\)
Vậy \(MAX_A=\dfrac{9}{5}\) khi \(x=\dfrac{-2}{5}\)
\(-5.\left(x^2+\dfrac{4x}{5}-\dfrac{1}{5}\right)\)\
\(\Leftrightarrow-5.\left(x^2+2.x.\dfrac{2}{5}+\dfrac{4}{25}-\dfrac{1}{5}-\dfrac{4}{25}\right)\)
\(\Leftrightarrow-5.\left(x+\dfrac{2}{5}\right)^2-\dfrac{9}{5}\le-\dfrac{9}{5}\)
Vậy Max = \(\dfrac{-9}{5}\)
Dấu "=" xảy ra khi x=\(\dfrac{-2}{5}\)
Đặt A= \(-5x^2-4x+1\)
Ta có :
\(-5x^2-4x+1=-5\left(x^2+\dfrac{4}{5}x+\dfrac{4}{25}\right)-\dfrac{21}{5}=-5\left(x+\dfrac{2}{5}\right)^2-\dfrac{21}{5}\le\dfrac{-21}{5}\)Vậy \(Max_A=\dfrac{-21}{5}\) khi \(x+\dfrac{2}{5}=0\Rightarrow x=\dfrac{-2}{5}\)
