初一 数学 因式分解 请详细解答,谢谢! (31 10:1:52)

初一 数学 因式分解 请详细解答,谢谢! (31 10:1:52)

1.a^2-b^2-c^2+2bc
=a^2-(b^2+c^2-2bc)
=a^2-(b-c)^2
=(a+b-c)(a-b+c)
2.x^2-y^2-x+1/4
=(x^2-x+1/4)-y^2
=(x-1/2)^2-y^2
=(x-1/2+y)(x-1/2-y)
3.由（x+2)^2-(y+2)^2=4得
（x+2+y+2)(x+2-y-2)=4
即（x+y+4)(x-y)=4
x,y互为相反数，则x+y=0 x-y=2x代入（x+y+4)(x-y)=4得
x=1/2 y=-1/2

1. =a^2-(b-c)^2=(a-b+c)(a+b-c)
2. =(x-1/2)^2-y^2=(x-1/2-y)(x-1/2+y)
3. (x+2)^2-(y+2)^2=(x-y)(x+y+4)＝4。y=-x，所以2x*4=4 => x=1/2 => y=-1/2

1.=a^2-（b-c)^2=(a+b-c)(a-b+c)
2=(x-.5)^2-y^2=(x-.5-y)(x-.5+y)
3=(x+2+y+2）(x+2-y-2)=4
有{x+y=0 x-y=1
x=.5 y=-.5

1 =a^2-(b-c)^2(完全平方)=(a+b-c)(a-b+c) (平方差)
2 =(x-1/2)^2-y^2(完全平方)=(x+y-1/2)(x-y-1/2)(平方差)
3 (x+2)^2-(-x+2)^2=4(2x)=4 x=1/2 y=-1/2

1. a^2-b^2-c^2+2bc=a^2-(b-c)^2=(a-b+c)(a+b-c)
2. x^2-y^2-x+1/4=(x-1/2)^2-y^2=(x-1/2-y)(x-1/2+y)
3. 因为(x+2)^2-(y+2)^2=(x-y)(x+y+4)＝4。又因为y=-x，
所以2x*4=4 => x=1/2 => y=-1/2

1.
a^2-b^2-c^2+2bc
=a^2-(b^2+c^2-2bc)
=a^2-(b-c)^2
=[a-(b-c)][a+(b-c)]
=(a-b+c)(a+b-c)
2.
x^2-y^2-x+1/4
=(x^2-x+1/4)-y^2
=(x-1/2)^2-y^2
=[(x-1/2)-y][(x-1/2)+y]
=(x-1/2-y)(x-1/2+y)
3.
根据题意得:
x+y=0
x=-y
(x+2)^2-(y+2)^2=4
[(x+2)-(y+2)][(x+2)+(y+2)]=4
(x+2-y-2)(x+2+y+2)=4
(x-y)(x+y+4)=4
4(x-y)=4
x-y=1
-y-y=1
-2y=1
y=-1/2
所以得:
x=1/2
y=-1/2