a) \(4x-x^2=-\left(x^2-4x+4\right)+4=-\left(x-2\right)^2+4\le4\forall x\)
- Vậy GTLN của biểu thức là 4, đạt tại \(x-2=0\Leftrightarrow x=2\).
b) \(-3x^2+5x-1=-3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)+3.\dfrac{25}{36}-1=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{13}{12}\le\dfrac{13}{12}\forall x\)
- Vậy GTLN của biểu thức là \(\dfrac{13}{12}\), đạt tại \(x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\).
a: \(=-\left(x^2-4x+4\right)+4=-\left(x-2\right)^2+4< =4\)
Dấu '=' xảy ra khi x=2
b: \(=-3\left(x^2-\dfrac{5}{3}x+\dfrac{1}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{13}{36}\right)\)
\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{13}{12}< =\dfrac{13}{12}\)
Dấu '=' xảy ra khi x=5/6
`a, = -x^2 + 4x - 4 + 4 = -(x-2)^2 + 4<=0 + 4 = 4`
`b, -3x^2 + 5x - 1 = -3(x^2 -5/3x + 1/3) = -3(x^2-5/3 + 25/36 - 13/36) = -3(x-5/6)^2 + 13/12 <=13/12`
a) \(4x-x^2\)
\(=-\left(x^2-4x+4\right)+4\)
\(=-\left(x-2\right)^2+4\)
\(Vì\) \(-\left(x-2\right)^2\le0\Rightarrow A\le4\)
\(MaxA=4\Leftrightarrow x=2\)
b) \(-3x^2+5x-1\)
\(=-3\left(x^2-2.x.\dfrac{5}{6}+\dfrac{25}{36}\right)+\dfrac{13}{12}\)
\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{13}{12}\)
\(Vì\) \(-3\left(x-\dfrac{5}{6}\right)^2\le0\Rightarrow B\le\dfrac{13}{12}\)
\(MaxB=\dfrac{13}{12}\Leftrightarrow x=\dfrac{5}{6}\)
