Question

因式分解1、(a+b)(a-b)+c(2b-c)2、已知4x^2+4xy+y^2-4x-2y+1=0,求证2x^2+3x

Answer 1

1、(a+b)(a-b)+c(2b-c)

=a^2-b^2+2bc-c^2

=a^2-(b-c)^2

=(a+b-b)(a-b+c)

2、4x^2+4xy+y^2-4x-2y+1=0

(2x+y)^2-2(2x+y)+1=0

(2x+y-1)^2=0

2x+y-1=0

2x+y=1

所以2x^2+3xy+y^2-x-y

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

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

=x+y-x-y

=0