a: Ta có: \(-x^2+2xy-4y^2+2x+10y-8\)

\(=-x^2+2xy-y^2+2x-2y-3y^2+12y-8\)

\(=-\left(x-y\right)^2+2\left(x-y\right)-1-3y^2+12y-12+5\)

\(=-\left(x-y-1\right)^2-3\left(y^2-4y+4\right)+5=-\left(x-y-1\right)^2-3\left(y-2\right)^2+5\le5\forall x,y\)

Dấu '=' xảy ra khi x-y-1=0 và y-2=0

=>y=2 và x=y+1=2+1=3

b: \(-x^2-y^2+xy+x+y\)

\(=-\frac14\left(4x^2+4y^2-4xy-4x-4y\right)\)

\(=-\frac14\left\lbrack4x^2-4xy+y^2-4x+2y+3y^2-6y\right\rbrack\)

\(=-\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2-6y+3-4\right\rbrack\)

\(=-\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y-1\right)^2-4\right\rbrack=-\frac14\left(2x-y-1\right)^2-\frac34\left(y-1\right)^2+1\le1\forall x,y\)

Dấu '=' xảy ra khi y-1=0 và 2x-y-1=0

=>y=1 và 2x=y+1=1+1=2

=>y=1 và x=1

A= -x²+2x+3

=>A= -(x²-2x+3)

=>A= -(x²-2.x.1+1+3-1)

=>A=-[(x-1)²+2]

=>A= -(x+1)²-2

Vì -(x+1)² ≤0=> A≤-2

Dấu "=" xảy ra khi

-(x+1)²=0 => x=-1

Vây A lớn nhất= -2 khi x= -1

B=x²-2x+4y²-4y+8

=> B= (x²-2x+1)+(4y²-4y+1)+6

=> B=(x-1)²+(2y+1)²+6

=> B lớn nhất=6 khi x=1 và y=-1/2

A=\(-x^2+2xy-4y^2+2x+8y-8=-\left(x^2-2xy+y^2-2x+1+2y\right)-\left(3y^2-6y+3\right)-4=-4-\left(x-y-1\right)^2-3\left(y-1\right)^2\le-4\)

=>Max A=-4<=>(x-y-1)²=0 và (y-1)²=0<=>x=2 y=1

nhìu giữ cha !!!!

a.

$12x^3y-24x^2y^2+12xy^3=12xy(x^2-2xy+y^2)=12xy(x-y)^2$
b.

$x^2-6x+xy-6y=(x^2+xy)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)$
c.

$2x^2+2xy-x-y=2x(x+y)-(x+y)=(x+y)(2x-1)$

d.

$x^3-3x^2+3x-1=(x-1)^3$

e.

$3x^2-3y^2-12x-12y=(3x^2-3y^2)-(12x+12y)$

$=3(x-y)(x+y)-12(x+y)=(x+y)[3(x-y)-12]=3(x-y)(x-y-4)$

f.

$x^2-2xy-x^2+4y^2=4y^2-2xy=2y(2y-x)$

C=(2x-1)(x-1)(2x^2-3x-1)+2017

=(2x^2-3x+1)(2x^2-3x-1)+2017

=(2x^2-3x)^2-1+2017

=(2x^2-3x)^2+2016>=2016

Dấu = xảy ra khi 2x^2-3x=0

=>x=0 hoặc x=3/2

D=(x-1)(x-6)(x-3)(x-4)+10

=(x^2-7x+6)(x^2-7x+12)+10

=(x^2-7x)^2+18*(x^2-7x)+72+10

=(x^2-7x+9)^2+1>=1

Dấu = xảy ra khi x^2-7x+9=0

=>\(x=\dfrac{7\pm\sqrt{13}}{2}\)

Ta có:

\(3-S=\left(x^2+4y^2+9z^2\right)-\left(2x+4y+6z\right)\)

\(\Leftrightarrow3-S=\left(x^2-2x+1\right)+\left(4y^2-4y+1\right)+\left(9z^2-6z+1\right)-3\)

\(\Leftrightarrow6-S=\left(x-1\right)^2+\left(2y-1\right)^2+\left(3z-1\right)^2\ge0\)

\(\Leftrightarrow S\le6\)

\(S_{max}=6\) khi \(\left\{{}\begin{matrix}x-1=0\\2y-1=0\\3z-1=0\end{matrix}\right.\) \(\Leftrightarrow\left(x;y;z\right)=\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\)

a) Áp dụng bất đẳng thức Cosi ta có :

\(x^2+1\geq 2x\\ 4y^2+1\geq 4y\\ 9z^2+1\geq 6z\)

Suy ra \(S\leq 6\)

Dấu = xảy ra khi \(x=1;y=\frac{1}{2}; z=\frac{1}{3}\)