恆等式，我唔識-- [複製鏈接]

引用第19樓swordgx於2006-12-10 00:13發表的:
我地本書淨係教(a+b)(a+c)=a(b+c)
而家要計埋呢D野- -

你本書教(a+b)(a+c)=a(b+c)?????
咁我勸你掉左本書佢=_=
定係你睇錯/打錯....?
a+b+a+c先等於a(b+c)
a(b+c)等於ab+ac
(a+b)(a+c)係等於a^2+ac+ab+bc

睇錯左了
A+B+A+C先係
眼訓精神差

pq(1-4r)-p^2(4r-1)
x+y^2+y+xy
呢兩條係點計？

引用第22樓swordgx於2006-12-10 13:09發表的:
pq(1-4r)-p^2(4r-1)
x+y^2+y+xy
呢兩條係點計？

pq(1-4r)-p^2(4r-1)
=pq(1-4r)+p^2(1-4r)
=(pq+p^2)(1-4r)
=p(p+q)(1-4r)

x+y^2+y+xy
=(x+xy)+(y+y^2)
=x(1+y)+y(1+y)
=(x+y)(1+y)

pq(1-4r)-p^2(4r-1)
=pq(1-4r)+p^2(1-4r)
=(pq+p^2)(1-4r)
=p(p+q)(1-4r)
呢個我唔明點計=[

引用第24樓swordgx於2006-12-10 14:13發表的:
pq(1-4r)-p^2(4r-1)
負負得正,-p^2(4r-1)=-4p^2r+p^2=p^2(1-4r)
=pq(1-4r)+p^2(1-4r)
抽common factor,兩個數都有(1-4r)
=(pq+p^2)(1-4r)
(pq+p^2)再抽common factor p
=p(p+q)(1-4r)
呢個我唔明點計=[

咁
2a(mn-n+kn)+(n-mn-kn)b
=(2a-b)(mn-n+kn)?

引用第26樓swordgx於2006-12-10 14:24發表的:
咁
2a(mn-n+kn)+(n-mn-kn)b
=(2a-b)(mn-n+kn)?

對啊....
點解你之前好似完全唔識咁既=_=
依家識就好

引用第27樓牙然於2006-12-10 14:42發表的:
對啊....
點解你之前好似完全唔識咁既=_=
依家識就好

老師淨係教過我書果個既a(b+c)
都冇教過呢D=[
5n(2x-5)^2-5(2x-5)^3
唔同次方要點抽common factor?><

引用第28樓swordgx於2006-12-10 14:45發表的:
老師淨係教過我書果個既a(b+c)
都冇教過呢D=[
5n(2x-5)^2-5(2x-5)^3
唔同次方要點抽common factor?><

5n(2x-5)^2-5(2x-5)^3
=5n(2x-5)^2-5(2x-5)(2x-5)^2
=5(2x-5)^2[n-2x-5)

你可以咁睇
x^3-x^2
=xxx-xx
=xx(x-1)
=x^2(x-1)

引用第29樓牙然於2006-12-10 14:51發表的:
5n(2x-5)^2-5(2x-5)^3
=5n(2x-5)^2-5(2x-5)(2x-5)^2
=5(2x-5)^2[n-2x-5)
.......

=5(2x-5)^2[n-2x-5)
點解會得出呢個答案=[?

引用第30樓swordgx於2006-12-10 14:57發表的:
=5(2x-5)^2[n-2x-5)
點解會得出呢個答案=[?

5n(2x-5)^2 - 5(2x-5)(2x-5)^2
兩個數都有5同埋(2x-5)
抽左出尼,變做
5(2x-5)^2 [n-(2x-5)]
明唔明?

引用第30樓swordgx於2006-12-10 14:57發表的:
=5(2x-5)^2[n-2x-5)
點解會得出呢個答案=[?

唔係...
佢打錯左個負姐...
ans應該係:

5(2x-5)^2(n-2x+5)

ok=]?

引用第32樓前於2006-12-10 15:39發表的:
ok=]?

本來想打括號的=_=
但變左一邊中括號...一邊小括號
未拆開-口-
sorry了

我死了.....
5(y-3x)-6a(3x+y)^2
(x-3y)^2-(x-y)(x-2y)
又唔識計，慘慘...

引用第34樓swordgx於2006-12-10 18:55發表的:
我死了.....
5(y-3x)-6a(3x+y)^2
(x-3y)^2-(x-y)(x-2y)
又唔識計，慘慘...

1)
5(y-3x)-6a(3x+y)^2
=-5(3x-y)-6a(3x-y)^2
=-(3x-y)[5-6a(3x-y)]
=-(3x-y)(5-18a-6ay)
第2題將佢拆散再計
費是打step
答案冇計錯應該係-y(3x-7y)

我想問下你地點樣打到[2次]??...

引用第35樓前於2006-12-10 23:09發表的:
1)
5(y-3x)-6a(3x+y)^2
=-5(3x-y)-6a(3x-y)^2
=-(3x-y)[5-6a(3x-y)]
.......

5(y-3x)-6a(3x+y)^2
=-5(3x-y)-6a(3x-y)^2<---無啦啦變左-6a(3x-y)既？姐係點？
=-(3x-y)[5-6a(3x-y)]

引用第37樓swordgx於2006-12-13 23:03發表的:
5(y-3x)-6a(3x+y)^2
=-5(3x-y)-6a(3x-y)^2<---無啦啦變左-6a(3x-y)既？姐係點？
=-(3x-y)[5-6a(3x-y)]

抽左(3x+y)出黎嘛

(a-2b)@