如a*a=a+1 , b*b=b+1,自a≠b,則a*a*a*a*a+b*b*b*b*b=________。
請列詳式並演算出答案為止,再寫出答案
奧數 如何解答
2013-10-18 3:04 am
回答 (5)
2013-10-18 6:21 am
✔ 最佳答案
As a^2 = a + 1 and b^2 = b + 1, so, a and b are the roots of the equation x^2 - x - 1 = 0.
ie. a + b = 1, ab = -1, therefore
a^5 + b^5
= (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4)
= (a + b)[(a + b)^4 - 5(a^3b + ab^3) - 5a^2b^2]
= (a + b){(a + b)^4 - 5ab[(a + b)^2 - 2ab] - 5(ab)^2}
= (1){1^4 - 5(-1)[1^2 - 2(-1)] - 5(-1)^2}
= 1 + 5(3) - 5
= 11
2013-10-28 4:12 am
a，b為x^2－x－1=0之兩根
a+b=1，ab=－1
x^2=x+1
a^2=a+1
b^2=b+1
a^2+b^2=(a+b)+2=3
x^3=x^2+x
a^3=a^2+a
b^3=b^2+b
a^3+b^3=(a^2+b^2)+(a+b)=3+1=4
x^4=x^3+x^2
a^4=a^3+a^2
b^4=b^3+b^2
a^4+b^4=(a^3+b^3)+(a^2+b^2)=4+3=7
x^5=x^4+x^3
a^5=a^4+a^3
b^5=b^4+b^3
a^5+b^5=(a^4+b^4)+(a^3+b^3)=7+4=11
a+b=1，ab=－1
x^2=x+1
a^2=a+1
b^2=b+1
a^2+b^2=(a+b)+2=3
x^3=x^2+x
a^3=a^2+a
b^3=b^2+b
a^3+b^3=(a^2+b^2)+(a+b)=3+1=4
x^4=x^3+x^2
a^4=a^3+a^2
b^4=b^3+b^2
a^4+b^4=(a^3+b^3)+(a^2+b^2)=4+3=7
x^5=x^4+x^3
a^5=a^4+a^3
b^5=b^4+b^3
a^5+b^5=(a^4+b^4)+(a^3+b^3)=7+4=11
2013-10-23 5:07 am
讚﹗根本5系常人諗到ge野……
2013-10-20 2:12 am
那些年是正確的～
但如果不想處理高次數的算式，可以由以上的繼續
5a + 5b + 6
= 5(a+b) + 6
= 5(1) + 6
= 11
但如果不想處理高次數的算式，可以由以上的繼續
5a + 5b + 6
= 5(a+b) + 6
= 5(1) + 6
= 11
2013-10-18 3:22 am
a*a = a + 1
a*a*a = (a + 1)*a = a*a + a = (a + 1) + a = 2a + 1
a*a*a*a = (2a + 1)*a = 2a*a + a = 2(a + 1) + a = 3a + 2
a*a*a*a*a = (3a + 2)*a = 3a*a + 2a = 3(a + 1) + 2a = 5a + 3
Similarly, b*b*b*b*b = 5b + 3
a*a*a*a*a + b*b*b*b*b
= (5a + 3) + (5b + 3)
= 5a + 5b + 6
2013-10-17 22:48:49 補充：
那些年是正確的～
但如果不想處理高次數的算式，可以由以上的繼續
5a + 5b + 6
= 5(a+b) + 6
= 5(1) + 6
= 11
a*a*a = (a + 1)*a = a*a + a = (a + 1) + a = 2a + 1
a*a*a*a = (2a + 1)*a = 2a*a + a = 2(a + 1) + a = 3a + 2
a*a*a*a*a = (3a + 2)*a = 3a*a + 2a = 3(a + 1) + 2a = 5a + 3
Similarly, b*b*b*b*b = 5b + 3
a*a*a*a*a + b*b*b*b*b
= (5a + 3) + (5b + 3)
= 5a + 5b + 6
2013-10-17 22:48:49 補充：
那些年是正確的～
但如果不想處理高次數的算式，可以由以上的繼續
5a + 5b + 6
= 5(a+b) + 6
= 5(1) + 6
= 11
收錄日期: 2021-04-20 14:26:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131017000051KK00174