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

\(Q=\left(x-y\right)^2-2\cdot\left(x-y\right)\cdot2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\)

\(Q=\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\)

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

\(Q=\left(-x-5y\right)^2\)

b) \(A=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)

\(A=\left(xy+2\right)^3-3\cdot2\cdot\left(xy+2\right)^2+3\cdot2^2\cdot\left(xy+2\right)-2^3\)

\(A=\left[\left(xy+2\right)-2\right]^3\)

\(A=\left(xy+2-2\right)^3\)

\(A=\left(xy\right)^3\)

\(A=x^3y^3\)

c) \(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(=\left(x^3+6x^2+12x+8\right)+\left(x^2-6x^2+12x-8\right)-\left(2x^3+24x\right)\)

\(=x^3+6x^2+12x+8+x^2-6x^2+12x-8-2x^3-24x\)

\(=\left(x^3+x^3-2x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x-24x\right)+\left(8-8\right)\)

\(=0\)

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

=(x-y-2x-4y)^2=(-x-5y)^2=x^2+10xy+25y^2

b: =(xy+2-2)^3=(xy)^3=x^3y^3

c: =x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x(x^2+12)

=24x+2x^3-2x^3-24x

=0

\(A=3x^3-6x^2+9x-3x^3+2x^2+5x^2-5x=x^2+4x\\ B=\left(x^2+xy+y^2\right)\left(x-y\right)=x^3-y^3\)

Bài 1:

\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)

Bài 2:

\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)

câu d của bài 2 là của bài 1 nha mình để nhầm chỗ huhu

a) \(\left(x^5+4x^3-6x^2\right):4x^2\)

\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)

\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)

b) 

Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)

c) (-2x⁵ : 2x²) + (3x² : 2x²) + (-4x^3 : 2x^2)

= \(-x^3+\dfrac{3}{2}-2x\)

d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)

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

\(=x-4\)

(dùng hẳng đẳng thức thứ 7)

Bài 2 :

a) 3x(x - 2) - 5x(1 - x) - 8(x² - 3)

= 3x² - 6x - 5x + 5x² - 8x² + 24

= (3x² + 5x² - 8x²) + (-6x - 5x) + 24 

= -11x + 24

b) (x - y)(x² + xy + y²) + 2y³

= x³ - y³ + 2y³

= x³ + y³ 

c) (x - y)² + (x + y)² - 2(x - y)(x + y)

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

= [(x - y) + x + y)² = [x - y + x + y] = (2x)² = 4x²

Bài 1 :

a]=  \(\frac{1}{4}\)x³ + x - \(\frac{3}{2}\).

b] => [x³ + x² -12 ] = [ x² +3 ][x-2] + [-6]

c]= -x³ -2x +\(\frac{3}{2}\).

d] = [ x³ - 64 ]  = [ x² + 4x + 16][ x- 4].

a) \(\left(2a-3\right)\left(a+1\right)-\left(a^2+6a+9\right):\left(a+3\right)\)

\(=\left(2a^2+2a-3a-3\right)-\left(a+3\right)^2:\left(a+3\right)\)

\(=2a^2-a-3-\left(a+3\right)\)

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

\(=2a^2-2a-6\)

b) \(\left(3x-5y\right)\left(-xy\right)^2-3x^2y^2+4x^2y^3\)

\(=\left(3x-5y\right)\cdot x^2y^2-3x^2y^2+4x^2y^3\)

\(=3x^3y^2-5x^2y^3-3x^2y^2+4x^2y^3\)

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

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

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

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

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

\(=4x-8\)

a) A = (x - 3)(x² + 3x + 9) - (x³ + 3)

= x³ - 3³ - x³ - 3

= (x³ - x³) + (-27 - 3)

= -30

b) B = (2x + 1)(4x² - 2x + 1) - 8(x + 1/2)(x² - 1/2 x + 1/4)

= (2x)³ + 1³ - 8[x³ + (1/2)³]

= 8x³ + 1 - 8(x³ + 1/8)

= 8x³ + 1 - 8x³ - 1

= (8x³ - 8x³) + (1 - 1)

= 0

\(a,\left(2a-3\right)\left(a+1\right)+\left(a^2+6a+9\right):\left(a+3\right)\\ =2a^2-a-3+\left(a+3\right)^2:\left(a+3\right)\\ =2a^2-a-3+a+3\\ =2a^2\\ b,\left(3x-5y\right)\left(-xy\right)^2-3x^2y^2+4x^2y^3\\ =3x^3y^2-5x^2y^3-3x^2y^2+4x^2y^3\\ =3x^3y^2-3x^2y^2-x^2y^3\\ c,x\left(x-2\right)^2-\left(x+2\right)\left(x^2-2x+4\right)+4x^2\\ =x^3-4x^2+4x-x^3-8+4x^2\\ =4x-8\)

a) Thực hiện phép nhân và hằng đẳng thức thu được

A =  x 3  – 2 x 2  – ( x 3 – 1 3 ); rút gọn A = 1 – 2 x 2 .

b) Đặt (xy – 1) làm nhân tử chung ta được B = 3(1 – xy).

Mình nghĩ là phân tích đa thức 

a)\(3x+2y+xy+6\)

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

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

b)\(2x^2+3xy-2y^2-10x-5y+12\)

\(=2x^2+\left(3y-10\right)x-\left(2y^2+5y-12\right)\)

\(=\left[2x+\left(y-4\right)\right]\left(x+2y+3\right)\)