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

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

A = \(4x^2\) - 8x + 3

= [\(\left(2x\right)^2\) - 2.2x.2 + \(2^2\)] \(-2^2\) + 3

= \(\left(2x-2\right)^2\) - 1

Ta có: \(\left(2x-2\right)^2\) ≤ 0 ∀ x

\(\left(2x-2\right)^2\) - 1 ≤ - 1

Hay A ≤ - 1

Dấu "=" xảy ra ↔ 2x - 2 = 0

2x = 2

x = 1

Vậy GTLN của A = - 1 ↔ x = 1

B = 6x \(-x^2\) + 2

= - (\(x^2\) - 6x) + 2

= - (\(x^2\) - 2.x.3 + \(3^2\)) \(-3^2\) + 2

= - \(\left(x-3\right)^2\) -7

Ta có: \(-\left(x-3\right)^2\) ≤ 0 ∀ x

\(-\left(x-3\right)^2\) - 7 ≤ - 7

Hay B ≤ - 7

Dấu "=" xảy ra ↔ - (x - 3) = 0

- x + 3 = 0

- x= - 3

x = 3

Vậy GTLN của B = - 7 ↔ x = 3

C = x(2 - 3x)

= 2x \(-3x^2\)

= - 3(\(x^2\) - \(\frac{3}{2}x\) )

= - 3(\(x^2\) - 2.x.\(\frac{3}{4}\) + \(\frac{3}{4}^2\)) \(-\frac{3}{4}^2\)

Ta có: \(-3\left(x+\frac{3}{4}\right)^2\) ≤ 0 ∀ x

\(-3\left(x+\frac{3}{4}\right)^2\) \(-\frac{9}{16}\) ≤ \(-\frac{9}{16}\)

Hay C ≤ \(-\frac{9}{16}\)

Dấu "=" xảy ra ↔ \(-3\left(x+\frac{3}{4}\right)\) = 0

- 3x \(-\frac{9}{4}\) = 0

- 3x = \(\frac{9}{4}\)

x = \(-\frac{3}{4}\)

Vậy GTLN của C = \(-\frac{9}{16}\) ↔ x = \(-\frac{3}{4}\)

a: \(-x^2+4x+1\)

\(=-\left(x^2-4x-1\right)\)

\(=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5\le5\)

Dấu '=' xảy ra khi x=2

b: \(x^2+2x+6=\left(x+1\right)^2+5\)

\(\Leftrightarrow\dfrac{2000}{\left(x+1\right)^2+5}\le400\)

Dấu '=' xảy ra khi x=-1

c: \(-9x^2+6x+19\)

\(=-\left(9x^2-6x-19\right)\)

\(=-\left(9x^2-6x+1-20\right)\)

\(=-\left(3x-1\right)^2+20\le20\)

Dấu '=' xảy ra khi x=1/3

d: \(=-\left(x^2+4x+y^2-2y\right)\)

\(=-\left(x^2+4x+4+y^2-2y+1-5\right)\)

\(=-\left(x+2\right)^2-\left(y-1\right)^2+5\le5\)

Dấu '=' xảy ra khi x=-2 và y=1

Bài 2:

\(A=-2x^2+3x-5\)

\(=-2\left(x^2+\frac{3x}{2}-\frac{5}{2}\right)\)

\(=-2\left(x^2-\frac{3x}{2}+\frac{9}{16}\right)-\frac{31}{8}\)

\(=-2\left(x-\frac{3}{4}\right)^2-\frac{31}{8}\le-\frac{31}{8}\)

Dấu = khi \(-2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)

Vậy \(Max_A=-\frac{31}{8}\Leftrightarrow x=\frac{3}{4}\)

Bài 1:

a)x²-4x²y+4xy

=x(x-4xy+y)

b)đề sai

Bài 3:

3yx + 6x - y = 7

<=> x(3y+6) - (3y+6) = 27

<=> (3y+6)(x+1) = 27

Ta có bảng sau:

Vậy...