數學問題(10分)

1)
w^3+12vw^2+36v^2w
=w(w^2+12vw+36v^2)
=w(w+6v)^2
2)
3(3a-b)^2=12k(b-3a)+12k^2
let (3a-b)=m
3m^2=12k(k-m)
m^2-4k^2+4km=0
(m^2+2k)^2=8k^2
m+2k=2√2k
m=2k(√2-1)
k = ((3a-b)(√2+1))/2
3)
(√a-1∕√a)^2
=(√(a^2) +1/(√a^2)-2(√a)(1∕√a)
=a+(1/a)-2

2007-07-27 08:29:57 補充：
3(3a-b)^2 12k(b-3a) 12k^2=3[(3a-b)^2-4k(3a-b) 4k^2]=3(3a-b-2k)^2

我抄人= = =a+(1/a)-2

w^3+2vw^2+36v^2w=???
= w^3+2vw^2+36v^2w
= 36v^2w+2vw^2+w^3
= w(36v^2+2vw+w^2)
= w[(6v)^2+2vw+w^2]

I wonder you have type the wrong question or not
I thing that the question should be
w^3+12vw^2+36v^2w=???
= w^3+12vw^2+36v^2w
= 36v^2w+12vw^2+w^3
= w(36v^2+12vw+w^2)
= w[(6v)^2+12vw+w^2]
= w(6v+w)^2



3(3a-b)^2=12k(b-3a)+12k^2=???
3(3a-b)^2=-12k(3a-b)+12k^2
3(3a-b)^2+12k(3a-b)- 12k^2=0
(3a-b)^2+4k(3a-b)- 4k^2=0
(3a-b + 2k) 2=0


展開(expand)
(√a-1∕√a)^2=???
=(√a-1)^2∕a
=(a - 2√a +1)∕a