Question

已知a+b+c=0,求a的平方-b的平方+ac-bc

Answer 1

1.a^2-b^2+ac-bc=(a-b)(a+b+c)=0

2.x^2+y^2-4x+6y+13=0

(x-2)^2+(y+3)^2=0

x=2 y=-3

2x+3y=-5

3,

（x+3）（x-2）+x^2-4

=(x-2)(2x+5)

4,

16x^4-7x^2*y^2+81y^4(不会化简,)

5,

3*(a^4+a^2*b^2+b^4)-9a^2*b^2

展开得：3*（a^4+2a^2*b^2+b^4）=3*（a^2+b^2）^2

6,

4x^2+2x-y^2-y

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

=(2x-y)*(2x+y+1)