Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

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

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

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

\(=-x^2-2x+8\)

\(=-\left(x^2+2x-8\right)\)

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

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

\(=9x^2-6xy+y^2+y^2+1\)

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)

a: =x^2-6x+9+x^2-6x+9

=2(x-3)^2>=0

Dấu = xảy ra khi x=3

b: =-(x^2+4x+y^2-2y)

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

=-(x+2)^2-(y-1)^2+5<=5

Dấu = xảy ra khi x=-2 và y=1

a: =-x^2+6x-4

=-(x^2-6x+4)

=-(x^2-6x+9-5)

=-(x-3)^2+5<=5

Dấu = xảy ra khi x=3

b: =3(x^2-5/3x+7/3)

=3(x^2-2*x*5/6+25/36+59/36)

=3(x-5/6)^2+59/12>=59/12

Dấu = xảy ra khi x=5/6

c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)

\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)

\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)

Dấu = xảy ra khi x=4 hoặc x=2

a: \(A=-x^2-4x-2\)

\(=-x^2-4x-4+2\)

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

\(=-\left(x+2\right)^2+2< =2\forall x\)

Dấu '=' xảy ra khi x+2=0

=>x=-2

b: \(B=-2x^2-3x+5\)

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

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}< =\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{3}{4}=0\)

=>\(x=-\dfrac{3}{4}\)

c: \(C=\left(2-x\right)\left(x+4\right)\)

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

\(=-x^2-2x+8\)

\(=-x^2-2x-1+9\)

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

\(=-\left(x+1\right)^2+9< =9\forall x\)

Dấu '=' xảy ra khi x+1=0

=>x=-1

d: \(D=-8x^2+4xy-y^2+3\)

\(=-8\left(x^2-\dfrac{1}{2}xy\right)-y^2+3\)

\(=-8\left(x^2-2\cdot x\cdot\dfrac{1}{4}y+\dfrac{1}{16}y^2\right)+\dfrac{1}{2}y^2-y^2+3\)

\(=-8\left(x-\dfrac{1}{4}y\right)^2-y^2+3< =3\forall x,y\)

Dấu '=' xảy ra khi y=0 và x-1/4y=0

=>y=0 và x=0

Bài 2: 

a: \(A=x^2+8x\)

\(=x^2+8x+16-16\)

\(=\left(x+4\right)^2-16\ge-16\)

Dấu '=' xảy ra khi x=-4

b: \(B=-2x^2+8x-15\)

\(=-2\left(x^2-4x+\dfrac{15}{2}\right)\)

\(=-2\left(x^2-4x+4+\dfrac{7}{2}\right)\)

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

Dấu '=' xảy ra khi x=2

c: \(C=x^2-4x+7\)

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

\(=\left(x-2\right)^2+3\ge3\)

Dấu '=' xảy ra khi x=2

e: \(E=x^2-6x+y^2-2y+12\)

\(=x^2-6x+9+y^2-2y+1+2\)

\(=\left(x-3\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu '=' xảy ra khi x=3 và y=1

\(A\left(x\right)=-\left(x^2-\frac{5}{3}x\right)+1=-3\left(x^2-2.x.\frac{5}{6}+\left(\frac{5}{6}\right)^2\right)+1+3.\left(\frac{5}{6}\right)^2\)

\(=-3\left(x-\frac{5}{6}\right)^2+\frac{37}{12}\le\frac{37}{12}\)

Dấu "=" xảy ra khi \(x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)

Vậy GTLN của A là 37/12.

b, c làm tương tự.