解F 2 Maths

👨🏻‍🏫 學術台數學

[匿名]

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

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!!

回答 (2)

用戶22403

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)

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╬跨次‧源﹏

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=

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