Question

初二数学因式分解(急需）

1.6x^2-x-15
2.x^2-4y^2+6x+9
3.x^2-4y^2-2x+4y
4.(x^2+5x+2)(x^2+5x+3)-12
5.m^2-12m-28
另附加：已知在三角形ABC中，三边A，B，C满足等式a^2-16b^2-c^2+6ab+10bc=0,求证a+c=2b
写过程，急需争取今天

Answer 1

1.6x^2-x-15
=6x^2-10x+9x-15
=2x(3x-5)+3(3x-5)
=(2x+3)(3x-5)

2.x^2-4y^2+6x+9
=(x^2+6x+9)-4y^2
=(x+3)^2-4y^2
=(x+3+2y)(x+3-2y)

3.x^2-4y^2-2x+4y
=(x+2y)(x-2y)-2(x-2y)
=(x-2y)(x+2y-2)

4.(x^2+5x+2)(x^2+5x+3)-12
=(x^2+5x)^2+5(x^2+5x)+6-12
=(x^2+5x)^2+5(x^2+5x)-6
=(x^2+5x+6)(x^2+5x-1)
=(x+2)(x+3)(x^2+5x-1)

5.m^2-12m-28
=(m-12)(m+2)

另附加：已知在三角形ABC中，三边A，B，C满足等式a^2-16b^2-c^2+6ab+10bc=0,求证a+c=2b
(a^2+6ab+9b^2)-(25b^2-10bc+c^2)=0
(a+3b)^2-(5b-c)^2
(a+3b+5b-c)(a+3b-5b+c)=0
(a+8b-c)(a-2b+c)=0
三角形两边之差小于第三边
c-a<b<8b
所以a+8b-c>0
所以a-2b+c=0
所以a+c=2b

Answer 2

1.（3X-5）（2X+3）
2.(X+3)~2-4Y~2=(X+3+2Y)(X+3-2Y)
3.(X-2Y)(X+2Y)-2(X-2Y)=(X-2Y)(X+2Y-2)
4.设x~2+5x=y，然后化简得到Y~2+5Y-6,所以(Y-6)(Y+1)即(X+6)(X-1)(X~2+5X+1)
5.(M-14)(M+2)

Answer 3

1.(2x+3)(3x-5)
2.(x+3+2y)(x+3-2y)
3.(x-2y)(x+2y-2)
4.(x+2)(x+3)(x^2+5x-1)
5.(m-12)(m+2)
6.a+8b-c>0
a-2b+c=0
a+c=2b

Answer 4

1)6x^2-x-15=(2x+3)(3x-5)
2)x^2-4y^2+6x+9=x^2-4y^2+6x+4+5=(x+1)(x+5)-(2y+3)(2y-3)
3)x^2-4y^2-2x+4y=x^2-2x+1-(4y^2-4y+1)=(x-1)^2-(2y-1)^2
4).(x^2+5x+2)(x^2+5x+3)-12=(x^2+5x)^2+5(x^2+5x)+6-12=(x^2+5x)^2+5(x^2+5x)-6=(x^2+5x-1)(x^2+5x+6)
5)m^2-12m-28=(m+2)(m-14)
二:a^2-16b^2-c^2+6ab+10bc=a^2+6ab+9b^2-(25b^2-10bc+c^2)=(a+3b)^2-(5b-c)^2=0
可知 a+3b=0 5b-c=0 ∴a+c=2b