Question

tim gtln cua
a=-4x^2-8x+3
b=6x-x^2+2
c=x(2-3x)
d=3x-x^2+2
e=3-2x^2+2xy-y^2-2x

Answer 1

a) \(A=-4x^2-8x+3=-4\left(x^2+2x+1\right)+7=-4\left(x+1\right)^2+7\le7\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x+1\right)^2=0\Rightarrow x=-1\)

Vậy Max(A) = 7 khi x = -1

b) \(B=6x-x^2+2=-\left(x^2-6x+9\right)+11=-\left(x-3\right)^2+11\le11\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-3\right)^2=0\Rightarrow x=3\)

Vậy Max(B) = 11 khi x = 3

c) \(C=x\left(2-3x\right)=-3\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+\frac{1}{3}=-3\left(x-\frac{1}{3}\right)^2+\frac{1}{3}\le\frac{1}{3}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-\frac{1}{3}\right)^2=0\Rightarrow x=\frac{1}{3}\)

Vậy Max(C) = 1/3 khi x = 1/3

Answer 2

d) \(D=3x-x^2+2=-\left(x^2-3x+\frac{9}{4}\right)+\frac{17}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)

Vậy Max(D) = 17/4 khi x = 3/2

e) \(E=3-2x^2+2xy-y^2-2x\)

\(E=-\left(x^2-2xy+y^2\right)-\left(x^2+2x+1\right)+4\)

\(E=-\left(x-y\right)^2-\left(x+1\right)^2+4\le4\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x+1\right)^2=0\end{cases}}\Rightarrow x=y=-1\)

Vậy Max(E) = 4 khi x = y = -1