(b)using the result of (a),prove that (y-3)y=(y-1)的2次-y-1 is also an identity
更新1:
thz!!!!!!!!!!!!!!!quick la...
更新2:
using the result of (a),wo~!!
更新3:
(b) (y-3)y=(y-1)^2-y-1 by (a)=y-1好似咁樣
更新4:
quick la..............急!!
解F 2 Maths
2010-10-28 4:21 am
(a) prove that (x-2)(x+1)=x的2次-x-2 is an identity
(b)using the result of (a),prove that (y-3)y=(y-1)的2次-y-1 is also an identity
(b)using the result of (a),prove that (y-3)y=(y-1)的2次-y-1 is also an identity
回答 (2)
2010-10-28 6:37 am
✔ 最佳答案
(a)LHS = (x-2)(x+1)
= x^2+x-2x-2
= x^2-x-2
= RHS
therefore, (x-2)(x+1) 三 x^2-x-2 (三 = 全等)
(b)
let y = x+1
LHS=(y-3)y
= (x+1-3)(x+1)
= (x-2)(x+1)
= x^2-x-2
RHS=(y-1)^2-y-1
=(x+1-1)^2-(x+1)-1
= x^2-x-1-1
= x^2-x-2
then,LHS=RHS
therefore, (y-3)y 三 (y-1)^2-y-1 (三 = 全等)
2010-10-28 4:38 am
(a)(x-2)(x+1)=x^2-x-2
LHS=(x-2)(x+1)
=x^2+x-2x-2 [x^2 = x的2次方]
=x^2-x-2
=RHS
Ans:(x-2)(x+1)≡x^2-x-2
(b)(y-3)y=(y-1)^2-y-1
LHS=(y-3)y
=y^2-3y [y^2 = y的2次方]
RHS=(y-1)^2-y-1
=y^2-2y+1-y-1
=y^2-3y
Ans:(y-3)y≡(y-1)^2-y-1
記住,最後Ans的地方 一定要用≡ 不可以用=
LHS=(x-2)(x+1)
=x^2+x-2x-2 [x^2 = x的2次方]
=x^2-x-2
=RHS
Ans:(x-2)(x+1)≡x^2-x-2
(b)(y-3)y=(y-1)^2-y-1
LHS=(y-3)y
=y^2-3y [y^2 = y的2次方]
RHS=(y-1)^2-y-1
=y^2-2y+1-y-1
=y^2-3y
Ans:(y-3)y≡(y-1)^2-y-1
記住,最後Ans的地方 一定要用≡ 不可以用=
收錄日期: 2021-04-23 23:17:50
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https://hk.answers.yahoo.com/question/index?qid=20101027000051KK01099