\(E=-x^2+4x+3=-\left(x-2\right)^2+7\le7\)
Dấu \("="\Leftrightarrow x=2\)
\(F=-8x+5-x^2=-\left(x+4\right)^2+21\le21\)
Dấu \("="\Leftrightarrow x=-4\)
\(K=-2x^2-4x+3=-2\left(x+1\right)^2+5\le5\)
Dấu \("="\Leftrightarrow x=-1\)
\(Q=2x-3-3x^2=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{8}{3}=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{8}{3}\le-\dfrac{8}{3}\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{3}\)
a, \(E=-x^2+4x+3=-\left(x^2-4x+4-4\right)+3=-\left(x-2\right)^2+7\le7\)
Dấu ''='' xảy ra khi x = 2
b, \(F=-x^2-8x+5=-\left(x^2+8x+16-16\right)+5=-\left(x+4\right)^2+21\le21\)
Dấu ''='' xảy ra khi x = -4
c, \(K=-2x^2-4x+3=-2\left(x^2+4x+4-4\right)+3=-2\left(x+2\right)^2+11\le11\)
Dấu ''='' xảy ra khi x = -2
d, \(Q=-3x^2+2x-3=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}-\dfrac{1}{9}\right)-3=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{8}{3}\le-\dfrac{8}{3}\)
Dấu ''='' xảy ra khi x = 1/3
a: Ta có; \(E=-x^2+4x+3\)
\(=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7\le7\forall x\)
Dấu '=' xảy ra khi x=2
