polynomial3
回答 (2)
2008-07-02 6:15 pm
This statement is false,
actually, you don't need to expand explicitly,
just check the coefficient of y^6, it's +1 at R.H.S, but -1 at L.H.S.
see?
(x+y)^5 中, y^5 的系數是+1, 乘 (x-y)後， y^6 系數是 -1, 與右側的 +1矛盾.
actually, you don't need to expand explicitly,
just check the coefficient of y^6, it's +1 at R.H.S, but -1 at L.H.S.
see?
(x+y)^5 中, y^5 的系數是+1, 乘 (x-y)後， y^6 系數是 -1, 與右側的 +1矛盾.
2008-07-02 3:28 am
no ,
LHS= (x+y)^5(x-y)
= ( x^5 +5x^4y+10x^3y^2+10x^2y^3+ 5xy^4 +y^5)( x-y) (二項式定理)
= x^6 + 4x^5y + ... +y^6 =/= x^6 + y^6 =RHS
LHS= (x+y)^5(x-y)
= ( x^5 +5x^4y+10x^3y^2+10x^2y^3+ 5xy^4 +y^5)( x-y) (二項式定理)
= x^6 + 4x^5y + ... +y^6 =/= x^6 + y^6 =RHS
收錄日期: 2021-04-19 01:22:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080701000051KK02309