square用^2去代替
(67)^2+b^2=c^2
c-b=6
求b同c
請問可唔可以求到個ans出黎
勾股定理問題,a^2+b^2=c^2 只俾a,冇b同c,但俾多個條件,c-b=6,可以答嗎
2007-01-09 2:02 am
回答 (3)
2007-01-09 2:12 am
✔ 最佳答案
67^2 + b^2 = c^267^2 = c^2 - b^2
67^2 = (c + b)(c - b)
So 67^2 = (c + b)(6)
c + b = 67^2 / 6 = 4489 / 6.............(1)
c - b = 6.........................(2)
(1) + (2), 2c = 4489 / 6 + 6
c = 4525 / 12
(1) - (2), 2b = 4489 / 6 - 6
b = 4453 / 12
2007-01-14 12:27 am
如果是普通一條勾股定理的問題, 那當然可以了.
估計題目不會超出普通數學的範圍, 而如果你有學過聯立方程的話, 若現有兩條式和總共不多於兩個未知數的話, 那就一定能解, 如無意外還會是唯一解.
問題是怎樣去把問題解得快一點.
估計題目不會超出普通數學的範圍, 而如果你有學過聯立方程的話, 若現有兩條式和總共不多於兩個未知數的話, 那就一定能解, 如無意外還會是唯一解.
問題是怎樣去把問題解得快一點.
2007-01-09 2:13 am
Simultaneous equations- 1 quadratic and 1 linear
(67)^2+b^2=c^2.... (1)
c-b=6.... (2)
From (2) c=6+b..... (3)
Sub (3) into (1) (67)^2+b^2= (6+b)^2
4489+b^2=36+12b+(b^2)
4489=36+12b
12b=4453
b=4453/12
Sub 4453/12 into (2) c-4453/12=6
c=4525/12
Terefore, b=4453/12, c=4525/12
(67)^2+b^2=c^2.... (1)
c-b=6.... (2)
From (2) c=6+b..... (3)
Sub (3) into (1) (67)^2+b^2= (6+b)^2
4489+b^2=36+12b+(b^2)
4489=36+12b
12b=4453
b=4453/12
Sub 4453/12 into (2) c-4453/12=6
c=4525/12
Terefore, b=4453/12, c=4525/12
參考: Me
收錄日期: 2021-04-13 13:46:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070108000051KK02856