Question

跪求几道数学因式分解题的答案。。。

1.x平方+y平方+2x-6y+10=0,求x+y=

2.x的3次方-2x平方y+xy方-x=

Answer 1

1.x^2+y^2+2x-6y+10=0;
即x^2+2x+1+y^2-6y+9=0
即(x+1)^2+(y-3)^2=0;
故x+1=0,y-3=0;
即x=-1,y=3
故x+y=2.

2.x^3-2x^2y+xy^2-x
=x(x^2-2xy+y^2)-x
=x(x-y)^2-x
=x[(x-y)^2-1]
=x(x-y+1)(x-y-1)

Answer 2

1，解：（x^2+2x+1)+(y^2-6y+9)=0
因为：(x+1)^2+(y-3)^2=0
所以：x+1=0,y-3=0
x=-1, y=3
把x=-1，y=3带入上式，
上式=-1+3=2=x+y

2，解：原式=x(x^2-2xy+y^2-1)
=x((x-y)^2-1)
=x(x-y-1)(x-y+1)