小明和小志步行12km。兩人以相同速率起步，但小明在步行2km後加快了1km/h，而小志步行速率維持不變。最後小明比小志快半小時完成步行，xkm/h為他們起步時速率。
(a)用x設一條方程
(b)求x
方程文字題 20點
2008-01-15 9:12 pm
回答 (4)
2008-01-16 5:40 am
（ａ） 12/x - [2/x + 10/(x+1)] = 1/2
(b) ＊12/x - (2/x + 10/(x+1)) = 1/2
12(x+1) - 2(x+1) - 10x = 1/2(x)(x+1)＊
＊...＊ -20 = x^2 + x
x = 4 or x =-5(rejected)
∴ x = 4
吾知系吾系呀...
(b) ＊12/x - (2/x + 10/(x+1)) = 1/2
12(x+1) - 2(x+1) - 10x = 1/2(x)(x+1)＊
＊...＊ -20 = x^2 + x
x = 4 or x =-5(rejected)
∴ x = 4
吾知系吾系呀...
參考: 自己...
2008-01-15 11:13 pm
a) 12/x - (2/x + 10/(x+1)) = 1/2
(b) 12/x - (2/x + 10/(x+1)) = 1/2
12(x+1) - 2(x+1) - 10x = 1/2(x)(x+1)
-20 = x^2 + x
x = 4 or x =-5(rejected)
so x = 4
(b) 12/x - (2/x + 10/(x+1)) = 1/2
12(x+1) - 2(x+1) - 10x = 1/2(x)(x+1)
-20 = x^2 + x
x = 4 or x =-5(rejected)
so x = 4
2008-01-15 10:00 pm
a) 12/x - (2/x + 10/(x+1)) = 1/2
(b) 12/x - (2/x + 10/(x+1)) = 1/2
12(x+1) - 2(x+1) - 10x = 1/2(x)(x+1)
-20 = x^2 + x
x = 4 or x =-5(rejected)
so x = 4
(b) 12/x - (2/x + 10/(x+1)) = 1/2
12(x+1) - 2(x+1) - 10x = 1/2(x)(x+1)
-20 = x^2 + x
x = 4 or x =-5(rejected)
so x = 4
2008-01-15 9:18 pm
自己功課自已做啦！
收錄日期: 2021-04-16 16:45:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080115000051KK01137