初中數學(20點!!)

2011-01-01 7:01 am
factorization
1. 2(2x+y)^2-6(2x+y)^3
2. (a-2b)^2+(2a-b)(2b-a)
3. 2(2a+b-1)^2-4(1-2a-b)
4. 5m^2n+m^2(1-n)+2m^2n
5. (5-2a)(5+a)-25-5a

factorization我比較差,希望各位可以比埋steps我,唔該晒
仲有2題其他類型既數唔係好識
1. 把公式h=2/h - 1/k的主項變換為k
2. 1^3, 2^3 ,3^3, 4^3, ...
以n表示該數列的第n項

再次唔該晒各位

回答 (2)

2011-01-01 7:39 am
✔ 最佳答案
2(2x+y)^2-6(2x+y)^3
= 2((2x+y)^2-3(2x+y)^3)
=2((2x+y)^2)(1-3(2x+y))
= 2((2x+y)^2)(1-6x-3y)

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

2(2a+b-1)^2-4(1-2a-b)
= 2(2a+b-1)((2a+b-1)-2)
= 2(2a+b-1)(2a+b-3)

5m^2n+m^2(1-n)+2m^2n cannot be factorized by simple methods =.=

(5-2a)(5+a)-25-5a
= (5-2a)(5+a) - 5(5+a)
= (5+a)(5-2a-5)
= -2a(5+a)

h=2/h - 1/k
(h^2)k = 2k - h (multiply the eq by hk)
(h^2 -2)k = -h
k=-h/(h^2 -2)

1^3, 2^3 ,3^3, 4^3, ...
for n=1, a_1=1^3
for n=2, a_2=2^3
for n=3, a_3=3^3
for n=4, a_4=4^3
so for any positive integers n, a_n = n^3
2011-01-02 1:56 am
HAHA
係唔係 math wb ch2f 呀
我都做緊呀
1. 2(2x+y)^2-6(2x+y)^3
=2(2x+y)^2[1-4(2x+y)]
=2(2x+y)(1-8x-4y)
2. (a-2b)^2+(2a-b)(2b-a)
=(a-2b)(a-2b)+(2a-b)(2b-a)
=(a-2b)(a-2b)+(2a-b)[-(a-2b)]
=(a-2b)(a-2b)-(2a-b)(a-2b)
=(a-2b)[a-2b(2a-b)]
=(a-2b)(a-2b-2a+b)
=(a-2b)(-a-b)
=-(a+b)(a-2b)
3. 2(2a+b-1)^2-4(1-2a-b)
=2(2a+b-1)^2-4[-(2b+b-1)]
=2(2a+b-1)^2+4(2b+b-1)
=2(2a+b-1)[2(2a+b-1)]
=2(2a+b-1)(4a+2b-2)


4. 5m^2n+m^2(1-n)+2m^2n
(我都未計到)
5. (5-2a)(5+a)-25-5a
=(5+a)(5-2a-5)
= -2a(5+a)


2011-01-02 13:52:06 補充:
5m^2n+m^2(1-n)+2m^2n
=5m^2n+m^2-m^2n+2m^2n
=(m^2n)(5-m^2+2)

你記得放番個n落黎


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