a: \(A=2\left(x+y\right)+3xy\left(x+y\right)+5x^2y^2\left(x+y\right)=0\)

b: \(B=3xy\left(x+y\right)+2x^2y\left(x+y\right)=0\)

\(\Rightarrow\)A=2(x+y)+3xy(x+y)+5x²y²(x+y)

Thay x+y=0 vào A

\(\Rightarrow\)A=0

2) Ta có:

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

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

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

Do \(x+y-2=0\Rightarrow x+y=2\)

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

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

\(=0+0+3\)

\(=3\)

Vậy \(B=3\)

1) Ta có:

\(A=x^3+x^2y-2x^2-xy-y^2+3y+x-1\)

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

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

\(=0+0+0+1\)

\(=1\)

Vậy \(A=1\)

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

     \(=1\)

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

Ta có : \(x+y-2=0\Rightarrow x+2=-y\)

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

     \(N=-xy\left(x+y-2\right)+2=2\)

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

A=2(x+y)+3xy(x+y)+5x²y²(x+y)+2

A=2.0+3xy.0+5x²y².0+2

A=2

B=xy(x+y)+2x²y (x+y)+5

B=xy.0+2x²y.0+5=5

a,Ta có 2(x+y)+3xy(x+y)+5x²y²(x+y)+4

Xg thay x+y=0 vào là dc bn nhó

Chúc bn hok tốt

Tui chẳng nghĩ gì về số cúp cả

trả lời đi t đag cần gấp lắm