The square of the sum of three positive consecutive numbers is greater than the sum of their squares by 94.Find these three positive consecutive numbers.
有冇人識計啊啊=-=...
中四數學題=.=!!
2008-09-30 2:50 am
回答 (2)
2008-09-30 3:07 am
✔ 最佳答案
Let the 3 numbers be x- 1, x and x + 1.Square of sum = ( x -1 + x + x +1)^2 = (3x)^2= 9x^2.
Sum of their square = (x -1)^2 + x^2 + (x + 1)^2
= x^2 + 1 - 2x + x^2 + x^2 + 2x + 1 = 3x^2 + 2.
So 9x^2 = 3x^2 + 2 + 94
6x^2 = 96
x^2 = 16
x = 4 for positive numbers.
So the 3 numbers are 3, 4 and 5.
2008-09-30 3:07 am
呢三個連續 numbers 係 3. 4. 5
( 3的平方 + 4的平方 + 5的平方 ) + 94 = 144
( 3 + 4 + 5 ) 的平方 = 144
2008-09-29 19:19:29 補充：
設三數為 x-1, x, x+1 , 符號^2 即平方, "切記 (x-1)^2 和 (x+1)^2是有公式"
(x-1)^2 + x^2 + (x+1)^2 + 94 = ( x-1+x+x+1) ^2
(x^2 - 2x + 1 + x^2 + x^2 + 2x + 1) + 94 = 9x^2
3x^2 + 2 + 94 = 9x^2
96 = 6x^2
16 = x^2
x = 4
故三個連續數為(4-1),4,(4+1) 即3,4,5
( 3的平方 + 4的平方 + 5的平方 ) + 94 = 144
( 3 + 4 + 5 ) 的平方 = 144
2008-09-29 19:19:29 補充：
設三數為 x-1, x, x+1 , 符號^2 即平方, "切記 (x-1)^2 和 (x+1)^2是有公式"
(x-1)^2 + x^2 + (x+1)^2 + 94 = ( x-1+x+x+1) ^2
(x^2 - 2x + 1 + x^2 + x^2 + 2x + 1) + 94 = 9x^2
3x^2 + 2 + 94 = 9x^2
96 = 6x^2
16 = x^2
x = 4
故三個連續數為(4-1),4,(4+1) 即3,4,5
參考: me
收錄日期: 2021-04-25 22:39:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080929000051KK01461