\(A=x^2-y^2+7x+7y\)

\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y+7\right)\)

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

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

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

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

\(=\left(x-3y\right)^2-9\)

\(=\left(x-3y-3\right)\left(x-3y+3\right)\)

A=\(x^2+7x+7y-y^2=\left(x^2-y^2\right)+\left(7x+7y\right)=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)=\left(x+y\right)\left(x-y+7\right)\)

B=\(4x^3-4x^2+x=x\left(4x^2-4x+1\right)=x\left(2x-1\right)^2\)

C=\(x^2+9y^2-9-6xy=\left(x^2-6xy+9y^2\right)-9=\left(x-3y\right)^2-3^2=\left(x-3y-3\right)\left(x-3y+3\right)\)

\(A=x^3-8-128-x^3=-136\\ B=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)

\(A=\left(x-2\right)\left(x^2+2x+4\right)-\left(128+x^3\right)=x^3-8-128-x^3=-136\)

\(B=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)-\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)

a) = (x - 4y)(x + 1)

b) = (x - 3y)^2 - 2^2

= (x - 3y - 2)(x - 3y + 2)

c) = x^2(x + 3) - 7x(x + 3) + 9(x + 3)

= (x + 3)(x^2 - 7x + 9)

a: \(x^2-4xy+x-4y\)

\(=x\left(x-4y\right)+\left(x-4y\right)\)

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

b: \(x^2-6xy+9y^2-4\)

\(=\left(x-3y\right)^2-4\)

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

2: \(H=2x^2+4y^2+4xy+4y+9\)

\(=2x^2+4xy+2y^2+2y^2+4y+2+7\)

\(=2\left(x+y\right)^2+2\left(y+1\right)^2+7\ge7\forall x,y\)

Dấu '=' xảy ra khi x+y=0 và y+1=0

=>y=-1 và x=-y=1

3: \(I=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2-4xy+4y^2+10x-20y+y^2-2y+28\)

\(=\left(x-2y\right)^2+10\left(x-2y\right)+25+y^2-2y+1+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi y-1=0 và x-2y+5=0

=>y=1 và x=2y-5=2*1-5=-3

4: \(K=x^2+5y^2-4xy+6x-14y+15\)

\(=x^2-4xy+4y^2+6x-12y+y^2-2y+15\)

\(=\left(x-2y\right)^2+6\left(x-2y\right)+9+y^2-2y+1+5\)

\(=\left(x-2y+3\right)^2+\left(y-1\right)^2+5\ge5\forall x,y\)

Dấu '=' xảy ra khi y-1=0 và x-2y+3=0

=>y=1 và x=2y-3=2*1-3=-1

b) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\dfrac{\left(2x+y\right)^2}{2x+y}=2x+y\)

c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\dfrac{\left(3y-x\right)^2}{3y-x}=3y-x\)

Lời giải:
$A=(x-3y)^2-15=[37-3(-1)]^2-15=40^2-15=1585$

a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)

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

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

$=4x^2-4x+1-2(4x^2-12x+9)+4$

$=4x^2-4x+1-8x^2+24x-18+4$

$=-4x^2+20x-13$

Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:

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

$=[(3x+2)-(2y-1)]^2$

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

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

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

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

Đặt $a=x-1,\ b=x$:

$=a^3+3a^2b+3ab^2+b^3$

$=(a+b)^3$

$=[(x-1)+x]^3$

$=(2x-1)^3$

Dùng $(a+b)(a^2-ab+b^2)=a^3+b^3$:

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

$=8x^3+27y^3$

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

$=-2y^3$

Dùng $(a-b)(a^2+ab+b^2)=a^3-b^3$:

$=(x^2)^3-(2y)^3-x^3(x^3-y^3)+8y^3$

$=x^6-8y^3-x^6+x^3y^3+8y^3$

$=x^3y^3$

a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)

a) $(2x-1)^2-2(2x-3)^2+4$

$=4x^2-4x+1-2(4x^2-12x+9)+4$

$=4x^2-4x+1-8x^2+24x-18+4$

$=-4x^2+20x-13$

b) $(3x+2)^2+2(2+3x)(1-2y)+(2y-1)^2$

Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:

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

$=[(3x+2)-(2y-1)]^2$

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

\(xy-3x-2y+6=x\left(y-3\right)-2\left(y-3\right)=\left(y-3\right)\left(x-2\right)\)

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