y(y+1)(x的平方+1）+x(2xy的平方+2y+1)

原式=x²y²+x²y+y²+y+2x²y²+2xy+x
=3x²y²+x²y+y²+y+2xy+xy(y+1)(x^2+1)+x(2y^2+2y+1) =y(y+1)(x^2+1)+x[y^2+(y^2+2y+1)] =x^2y(y+1)+y(y+1)+xy^2+x(y+1)^2 =xy[x(y+1)+y]+(y+1)[y+x(y+1)] =xy(xy+x+y)+(y+1)(xy+x+y) =(xy+x+y)(xy+y+1)这个对，还是您这个对呀？都对我是计算，你的是因式分解采纳吧