盡量幫幫我！有詳細過程
1) 根據: (a+1)^2=a^2+2a+1
expand: (x+1+y)^2
2) 根據: (x-1)^2=x^2-2x+1, (x+1)^2=x^2+2x+1
show that: (x^2-1)^2=(x^2+1)^2-4x^2
3) prove that it is an identity: (2x+y)/2- (6x+y)/3+ (4x+x)/4=5/y
4) 根據：(x+3)(x-3)=x^2-9
show that (y+1)(y-5)=(y-2)^2-9 it is an identity
5)如果： x-1/x=7
x^2+1/x^2=?
expandtion&identity? 急！救命！（10）very easy
2008-05-01 11:58 pm
回答 (1)
2008-05-02 5:31 pm
✔ 最佳答案
1. (x+1+y)^2=[(x+y)+1]^2=[(x+y)^2+2(x+y)+1]
=x^2+2xy+y^2+2x+2y+1
2. (x^2-1)^2=(x+1)(x-1)
(x^2+1)^2-4x^2=(x^2+1)^2-(2x)^2
=(x^2+1+2x)(x^2+1-2x)
=(x+1)(x-1)
therefore, they are equal
3. (2x+y)/2-(6x+y)/3+4x+y)/4
= (12x+6y-24x-4y+12+3y)12
=5y/12
4. (y-2)^2-9=(y-2)^2-3^2
=(y-2+3)(y-2-3)
=(y+1)(y-5)
5. (x-1/x)=7
(x-1/x)^2=49
x^2-2+1/x^2=49
x^2+1/x^2=51
收錄日期: 2021-04-29 20:25:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080501000051KK01925