問幾題數學題~10!!!!!!!!!!!!!!!!

用恆等式作因式分解 好易......................
因 應用式a(a+b)²=a²+2ab+b²
應用式b(a-b)²=a²-2ab+b²
應用式c(a+b)(a-b)=a²-b²



1. (a+b)² +4(a+b)+4 ----------------------------ans:(a+b+2)²
=(a+b)² + 2(2)(a+b) +2² 應用式a
=(a+b+2)²

2.(x-3y)² +6(x+3y)+9 ----------------------------ans:(x+3y+3)²
=(x+3y)²+2(3)(x+3y)+3²應用式b
=(x+3y+3)²

3.x² -y² +6x-2y ----------------------------ans:(x+y)(x-y+4)
=我做唔到因式分解@@你係唔係打錯式???

4.9x² -y² +6x-2y ----------------------------ans:(3x-y)(3x+y+2)
=(3x-y)(3x+y) +2(2x-y) 應用式c
=(3x-y)(3x+y+2)

5. (x² +4x+4)-9y² ----------------------------ans:(x+2-3y)(x+2+3y)
=(x²+2)² -(3y)² 應用式a+應用式c
=(x+2-3y)(x+2+3y)

6. (x² +6xy+9y² )-16z² ----------------------------ans:(x+3y-4z)(x+3y+4z)
=(x+3y)² -(4z)² 應用式a+應用式c
=:(x+3y-4z)(x+3y+4z)

2006-12-04 18:43:44 補充：
睇唔到第7題添 kuso!!!7. 1001² ans 1002001 =就咁開方吧啦@@=(1001)(1001)=1002001

2006-12-04 18:45:39 補充：
re:3.x² -y² 6x-2y ----------------------------ans:(x y)(x-y 4)=我做唔到因式分解@@你係唔係打錯式???如果個答案冇錯咁條式就係 x² -y² 4x-4y 如果個問題冇錯咁答案就係就係 不能因式分解