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標題: 唔識計@___________@ [打印本頁]

作者: sing1993 時間: 30/10/2005 11:01 AM
標題: 唔識計@___________@
x>0
x(y+z)=5
y(x+z)=10
z(x+y)=13
X=??????????

作者: Royal 時間: 30/10/2005 11:16 AM
下?= =淨係比左個"x>0"比你咋?!
根本就計唔到= =

作者: Kimo 時間: 30/10/2005 11:22 AM
xy+xz=5
xy+yz=10
xz+yz=13

yz-xz=5
xz+yz=13

yz=5+xz
xz+yz=13

xz+5+xz=13
2xz=8
xz=4

xy+4=5
xy+yz=10
4+yz=13

xy=1
xy+yz=10
yz=9

(xyz)^2=36
   xyz=6

xz=4
xy=1
yz=9
xyz=6

xyz/yz=6/9
   x=2/3

xyz/xz=6/4
   y=3/2

xyz/xy=6/1
   z=6

唔知o岩唔o岩 Orz

[ Last edited by Kimo on 2005-10-30 at 11:32 AM ]

作者: 綿羊 時間: 30/10/2005 11:45 AM
我竟然唔識計

作者: 鴨 時間: 30/10/2005 03:10 PM

Originally posted by Kimo at 2005-10-30 11:22 AM:
.....
(xyz)^2=36
xyz=6

...

呢道完全唔明.....

作者: Kimo 時間: 30/10/2005 03:44 PM

Originally posted by RX78-2 at 30-10-2005 15:10:

呢道完全唔明.....

xz=4
xy=1
yz=9

xxyyzz=4(9)
so (xyz)^2=36
√(xyz)^2=√36

作者: 路人2號 時間: 30/10/2005 05:03 PM

Originally posted by sing1993 at 2005-10-30 11:01 AM:
x>0
x(y+z)=5
y(x+z)=10
z(x+y)=13
X=??????????

你們教的真快.....
我們現在才做到9b.....T.Tlll

作者: Kimo 時間: 30/10/2005 05:09 PM

Originally posted by 路人2號 at 30-10-2005 17:03:

你們教的真快.....
我們現在才做到9b.....T.Tlll

我也只是F.1而已...樓主跟我同年的
不過唔知點解我maths特別勁，P.2開始年年maths全級第一(晒

作者: Echo21 時間: 30/10/2005 06:11 PM

Originally posted by Kimo at 2005-10-30 11:22:
...

你這樣做小心被扣Presentation分......
少做這類只有一個煩字的數......

x>0
x(y+z)=5
y(x+z)=10
z(x+y)=13

xy + xz = 5 ----------(1)
xy + yz = 10 ----------(2)
xz + yz = 13 ----------(3)

From (3),
yz = 13 - xz ----------(4)

Sub. (4) into (2)
xy - xz = -3
xy = -3 + xz ----------(5)

From (1),
xy = 5 - xz ----------(6)

Sub. (6) into (5)
5 - xz = -3 + xz
2xz = 8
xz = 4 ----------(7)

From (1), (4), (5), we have
xy = -3 + 4 = 1
yz = 13 - 4 = 9 ----------(8)

xy．xz．yz = 1．4．9
(xyz)^2 = 36
xyz = 6 ----------(9)

(9) / (8):
x = 2 / 3//

作者: 幻藍 時間: 30/10/2005 06:47 PM
只是比較複雜的代入問題吧......

作者: 鴨 時間: 30/10/2005 07:38 PM

Originally posted by Kimo at 2005-10-30 05:09 PM:

我也只是F.1而已...樓主跟我同年的
不過唔知點解我maths特別勁，P.2開始年年maths全級第一(晒

知你叻啦...唔駛係度曬....

Originally posted by 可愛的兔兒 at 2005-10-30 06:47 PM:
只是比較複雜的代入問題吧......

死囉...我F.3都唔係好明....

作者: sing1993 時間: 7/11/2005 07:32 PM

Originally posted by 路人2號 at 2005-10-30 17:03:

你們教的真快.....
我們現在才做到9b.....T.Tlll

nono.....上次咩鬼測驗果d...=.=

作者: 路人2號 時間: 9/11/2005 09:20 PM

Originally posted by sing1993 at 2005-11-7 07:32 PM:

nono.....上次咩鬼測驗果d...=.=

我們班測驗好像沒有測到這些膠東東喔xddd.....

作者: 碧咸 時間: 10/11/2005 08:27 PM
三條equation
三個unknown
要solve實得

作者: 宇文水靈 時間: 13/11/2005 10:30 PM
No need to be so trouble. I have an easy way to do!!!
x>0
x(y+z)=5
y(x+z)=10
z(x+y)=13
xy + xz = 5
xy + yz = 10
xz + yz = 13
Let a=xy, b=yz, c=xz.
a+c=5----------------------------(1)
a+b=10--------------------------(2)
b+c=13--------------------------(3)
(2) - (1) b-c=5----------------(4)
(3) + (4)  2b=18
            b=9
From equation (3), c=4
From equation (1), a=1
We have xy=1, --------------------(5)
         yz=9, --------------------(6)
         xz=4  --------------------(7)
From equation (5), x=1/y------(8)
Put (8) into (7), z/y=4
                     z=4y----------(9)
Put (9) into (6), 4y^2=9
                           y= +-3/2---(10)
Put (10) into (8), x= +-2/3
Since x > 0, x= -2/3 and y= -3/2 have to be rejected.
Therefore, x=2/3 and y=3/2
Put y = 3/2 into (9), z=6
Overall, x=2/3
         y=3/2
         z=6

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