飛航模式

此文章來自奇摩知識+如有不便請留言告知

標題:

F3 Maths Problems

發問:

The mean of x, y and z is 18 and the mean of x and y is 15. (a) Find the value of z. (b) It is also given that the mean of xy, yz and zx is 312. (1)Show that xy=216. (2) If x>y, find the value of x. Need Steps, Plz!

最佳解答:

(x + y + z)/3 = 18 ? x + y + z = 18 * 3 = 54 ...[1] (x + y)/2 = 15 ? x + y = 15 * 2 = 30 ...[2] (a) [1] - [2] gives z = 54 - 30 = 24 (b) (1) (xy + yz + zx)/3 = 312 ? xy + yz + zx = 312 * 3 = 936 ? xy + (y + x)z = 936 ? xy + (30)(24) = 936 by [2] and (a) ? xy + 720 = 936 ? xy = 216 (2) Consider {x + y = 30 {xy = 216 x(30 - x) = 216 30x - x2 = 216 x2 - 30x + 216 = 0 (x - 12)(x - 18) = 0 x = 12 or x = 18 (x, y) = (12, 18) or (18, 12) Since x > y, x = 18. (and y = 12)

其他解答:

1. x+y+z/3 =18 x+y+z=54 (18x3) 2. x+y/2=15 x+y=30(15x2) (a)Therefore x+y+z=54 30+z=54 z=24 b1 xy+yz+zx/3=312 xy+yz+zx=936 (312x3) xy+(y+x)z=936 xy+(30)24=936 xy+720=936 xy=216 (936-720) b2....只知道x=18 y=12..... P.S.Form1仔1個,都係將啲點讓比人啦=='A9A39959EDCC7BE3