a) 5xy² . (-3y)²

= 5xy² . 9y²

= (5.9).x.(y².y²)

= 45xy⁴

Hệ số: 45

Bậc: 5

b) x²yz . (-2xy)³

= x²yz . (-8x³y³)

= -8.(x².x³).(y.y³).z

= -8x⁵y⁴z

Hệ số: -8

Bậc: 10

c) (-2x²y)².8x³yz³

= 4x⁴y².8x³yz³

= (4.8).(x⁴.x³).(y².y).z³

= 32x⁷y³z³

Hệ số: 32

Bậc: 13

d) (-2xy³)².(-2xyz)³

= 4x²y⁶.(-8x³y³z³)

= [4.(-8)].(x².x³).(y⁶.y³).z³

= -32x⁵y⁹z³

Hệ số: -32

Bậc: 17

e) (-5xy³z).(-4x²)²

= (-5xy³z).(16x⁴)

= (-5.16).(x.x⁴).y³.z

= -80x⁵y³z

Hệ số: -80

Bậc: 9

f) (2x²y³)².(-2xy)

= (4x⁴y⁶).(-2xy)

= [4.(-2)].(x⁴.x).(y⁶.y)

= -8x⁵y⁷

Hệ số: -8

Bậc: 12

a: =5xy^2*9y^2=45xy^4

b: =x^2yz*(-8)x^3y^3=-8x^5y^4z

c: =4x^4y^2*8x^3yz^3=32x^7y^3z^3

d: =4x^2y^6*(-8)x^3y^3z^3=-32x^5y^9z^3

e: =-5xy^3z*16x^4=-80x^5y^3z

f: =4x^4y^6*(-2xy)=-8x^5y^7

a.16/4 . x = 16/2

x = 16/2 : 16/4

x = 16/2 . 4/16

x = 1/2.

b, b.2xy³+ 4x²y+8xy³−1x²y

= 2xy³ + 8xy³ + 4x²y - 1x²y

= ( 2 + 8 )xy³ + ( 4 - 1 )x²y

= 10xy³ + 3x²y.

a: x=16/2:16/4=2

b: \(=10xy^3+3x^2y\)

c: \(15x-14x+3xyz^3+2xyz^3-4xy^2+xy^2\)

\(=x+5xyz^3-3xy^2\)

Bài 1:

a: \(7xy^2\left(9xy^4+2xy+1\right)\)

\(=7xy^2\cdot9xy^4+7xy^2\cdot2xy+7xy^2\cdot1\)

\(=63x^2y^6+14x^2y^3+7xy^2\)

b: \(\frac12xyz\left(6x^2y^3+4x^3y-3xy\right)\)

\(=\frac12xyz\cdot6x^2y^3+\frac12xyz\cdot4x^3y-\frac12xyz\cdot3xy\)

\(=3x^3y^4z+2x^4y^2z-\frac32x^2y^2z\)

Bài 2:

1: \(5x^2y+10xy^3=5xy\cdot x+5xy\cdot2y^2=5xy\left(x+2y^2\right)\)

2: \(15xy^4-9xy=3xy\cdot5y^3-3xy\cdot3=3xy\left(5y^3-3\right)\)

3: \(5x^2+10xy+5y^2=5\left(x^2+2xy+y^2\right)=5\left(x+y\right)^2\)

Ta có: \(\left(3x^3-2xyz+\frac12xy^2\right)\cdot\left(-2xy\right)\)

\(=-3x^3\cdot2xy+2xyz\cdot2xy-\frac12xy^2\cdot2xy\)

\(=-6x^4y+4x^2y^2z-x^2y^3\)

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

b) \(x^3-4x^2-9x+36\\ =x^2\cdot x-4x^2-9x+36\\ =x^2\left(x-4\right)-9\left(x-4\right)\\=\left(x-4\right)\left(x^2-9\right)\\ =\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

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

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

e) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\\ =x^2y+xy^2+xyz+x^2z+xz^2+xyz+y^2z+yz^2\\ =xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy+xz+yz\right)\\ =\left(y+z\right)\left(z+x\right)\left(x+y\right)\)

Câu a kq lầ (x-y)(2+x+y) chứ

có 2xy +x +y = 7

(2xy + x)+y = 7

x. (2+y)+1.(2+y)=9

(2+y) . (x+1) = 9

Mà x;y E Z =>2+y ; x+1 E Z

                  =>2+y ; x+1 E ư (9)={1 ; -1 ; 3 ; -3 ; 9 ; -9}

BGT

vậy (x;y)=(0;1) ; (-2;-5) ; (2;-1) ; (-4;-3) ; (0;7) ; (-2;-11) ; (8;-1) ; (-10;-3) ; (2;7) ; (-4;-11)

mik là ng trả lời đầu tiên nên cũng ko chắc lắm nhé bn :>>

2xy + x + y = 7

x(2y + 1) + y = 7

2.[x(2y +1) + y ] = 2.7

2x(2y + 1) + 2y = 14

2x(2y+1) + 2y + 1 = 14 +1

2x(2y+1) + (2y +1) = 15

(2y+1).(2x+1)   = 15

Vì x, y thuộc Z nên 2x+1 và 2y+1 là ước của 15

*(mình làm đến đây bạn tự kẻ bảng nhé)

a) \(2xy-y+6x-3=\left(2xy+6x\right)-\left(y+3\right)=2x\left(y+3\right)-\left(y+3\right)=\left(2x-1\right)\left(y+3\right)\)

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

a) x^3 - 2xy^2 + 4y^3

b) x^2 + 3

chúc bạn hc tốt