B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

Thay x=2, y=-1/2 vào B ta có:
\(B=x^3+2x^2y-4xy^2+2y-3\\=2^3+2.2^2.\left(-\dfrac{1}{2}\right)-4.2.\left(-\dfrac{1}{2}\right)^2+2.\left(-\dfrac{1}{2}\right)-3\\ =8-4-2-1-3\\ =-2\)

Thay x = 2 ; y = -1/2 ta được 

\(B=8+2.4\left(-\dfrac{1}{2}\right)-\dfrac{4.2.1}{4}+2\left(-\dfrac{1}{2}\right)-3\)

\(=8-4-2-1-3=-2\)

( x + 2y )² - ( 2x - y ) ( x + 2y ) 

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

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

Tại x = 3 

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

Theo mik là vậy nha :D

Hình như bạn thiếu y nha bn

Biến đổi mỗi đa thức theo hướng làm xuất hiện thừa số x+y-2 \(M=x^3+x^2y-2x^2-xy-y^2+3y+x-1\)

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

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

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

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

\(M=x^2.0+y.0+0+1\)

\(M=1\)

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

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

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

\(N=\left(x^2x+x^2y-2x^2\right)-\left(xyx+xyy-2xy\right)+\left(2x+2y-4\right)+2\)

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

\(N=x^2.0-xy.0+2.0+2\)

\(N=2\)

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

\(P=\left(x^4+x^3y-2x^3\right)+\left(x^3y+x^2y^2-2x^2y\right)-\left(x^2+xy-2x\right)+3\)\(P=\left(x^3x+x^3y-2x^3\right)+\left(x^2y.x+x^2yy-2x^2y\right)-\left(xx+xy-2x\right)+3\)

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

\(P=x^3.0+x^2y.0-x.0+3\)

\(P=3\)

Tích mình nha!

Mà bài này hình như học ở lớp 7 rồi!

Biến đổi mỗi đa thức theo hướng làm xuất hiện thừa số x+y-2 M=x3+x2y−2x2−xy−y2+3y+x−1M=x3+x2y−2x2−xy−y2+3y+x−1

M=x3+x2y−2x2−xy−y2+(2y+y)+x−(−2+1)M=x3+x2y−2x2−xy−y2+(2y+y)+x−(−2+1)

M=(x3+x2y−2x2)−(xy+y2−2y)+(x+y−2)+1M=(x3+x2y−2x2)−(xy+y2−2y)+(x+y−2)+1

M=(x2.x+x2.y−2x2)−(x.y+y.y−2y)+(x+y−2)+1M=(x2.x+x2.y−2x2)−(x.y+y.y−2y)+(x+y−2)+1

M=x2.(x+y−2)−y.(x+y−2)+(x+y−2)+1M=x2.(x+y−2)−y.(x+y−2)+(x+y−2)+1

M=x2.0+y.0+0+1M=x2.0+y.0+0+1

M=1M=1

N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−2N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−2

N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−(−4+2)N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−(−4+2)

N=(x3+x2y−2x2)−(x2y+xy2−2xy)+(2x+2y−4)+2N=(x3+x2y−2x2)−(x2y+xy2−2xy)+(2x+2y−4)+2

N=(x2x+x2y−2x2)−(xyx+xyy−2xy)+(2x+2y−4)+2N=(x2x+x2y−2x2)−(xyx+xyy−2xy)+(2x+2y−4)+2

N=x2(x+y−2)−xy(x+y−2)+2(x+y−2)+2N=x2(x+y−2)−xy(x+y−2)+2(x+y−2)+2

N=x2.0−xy.0+2.0+2N=x2.0−xy.0+2.0+2

N=2N=2

P=x4+2x3y−2x3+x2y2−2x2y−x(x+y)+2x+3P=x4+2x3y−2x3+x2y2−2x2y−x(x+y)+2x+3

P=(x4+x3y−2x3)+(x3y+x2y2−2x2y)−(x2+xy−2x)+3P=(x4+x3y−2x3)+(x3y+x2y2−2x2y)−(x2+xy−2x)+3P=(x3x+x3y−2x3)+(x2y.x+x2yy−2x2y)−(xx+xy−2x)+3P=(x3x+x3y−2x3)+(x2y.x+x2yy−2x2y)−(xx+xy−2x)+3

P=x3(x+y−2)+x2y(x+y−2)−x(x+y−2)+3P=x3(x+y−2)+x2y(x+y−2)−x(x+y−2)+3

P=x3.0+x2y.0−x.0+3P=x3.0+x2y.0−x.0+3

P=3

-2 mới đúng nha

mình tưởng mũ 2y

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

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

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

Do \(x+y-2=0\Rightarrow (x+y-2)(x^3+x^2y-x)=0\)

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

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

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

Do \(x+y-2=0\Rightarrow\left(x+y-2\right)\left(x^3+x^2y-x\right)=0\)

k nhé 

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

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

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

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

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

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

\(A=0\)