\(P=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)

\(P=x^2\left(x+y\right)-2x^2-xy-y^2+3y+x+2017\)

\(P=2x^2-2x^2-xy-y^2+3y+x+2017\)

\(P=-xy-y^2+3y+x+2017\)

\(P=-y\left(x+y\right)+3y+x+2017\)

\(P=-2y+3y+x+2017\)

\(P=x+y+2017\)

\(P=2+2017=2019\)

P=x²(x+y)-2x²-y(y+x)+3y+x+2017

P=2x²-2x²-2y+3y+x+2017

P=2019

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!

Lời giải:

Ta có:

\(P=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)

\(P=x^2(x+y-2)-y(x+y)+3y+x+2017\)

\(P=x^2(x+y-2)-y(x+y)+(x+y)+2y+2017\)

\(P=x^2(2-2)-2y+2+2y+2017\)

\(P=2019\)

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

     \(=1\)

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

Ta có : \(x+y-2=0\Rightarrow x+2=-y\)

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

     \(N=-xy\left(x+y-2\right)+2=2\)

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