數學問題......請回答

2010-08-21 11:04 pm

係關於identities

If (x+1)(x+1)+A(2x+3)=x(x)+2Bx+4B
find the values of A and B.

我唔識打2次方果D..

P.S.請用英文作答....
THZ

更新1:

重有一題.. Expamd(a-b)(a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16)............(a^1024+b^1024). Briefly explain how you get the answer.. 如上

回答 (1)

STEVIE-G™

2010-08-21 11:07 pm

✔ 最佳答案

(x+1)^2+A(2x+3)=x^2+2Bx+4B
(x^2+2x+1)+(2Ax+3A)=x^2+2Bx+4B
x^2+(2+2A)x+(1+3A)=x^2+2Bx+4B

Comparing the coefficients, we have:
2+2A=2B => 1+A=B...(1)
1+3A=4B...(2)

Sub (1) into (2),
1+3A=4(1+A)
1+3A=4+4A
A=-3 and B=1-3=-2

2010-08-21 16:24:49 補充：
(a-b)(a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16)............(a^1024+b^1024)
=(a^2-b^2)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16)............(a^1024+b^1024)
=(a^4-b^4)(a^4+b^4)(a^8+b^8)(a^16+b^16)............(a^1024+b^1024)
=....
=(a^1024-b^1024)(a^1024+b^1024)
=a^2048-b^2048

2010-08-21 16:25:51 補充：
Get the answer by using the identity a^2-b^2=(a+b)(a-b).
同埋唔好係補充再發問新問題,今次我答埋你...

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