請給過程
(x+1):(y+2):z-3=2:3:4,且x+y+z=18則:
(1)x=?
(2)(2x+y):(z+5)=?
國1連比數學
2011-05-06 7:06 am
回答 (3)
2011-05-06 3:29 pm
✔ 最佳答案
(x+1):(y+2):z-3=2:3:4,且x+y+z=18則:假設 (x+1)/2 = (y+2)/3 = (z-3)/4 = p
x = 2p-1,
y = 3p-2,
z = 4p+3,
且 x+y+z=18
得 (2p-1) + (3p-2) + (4p+3) = 18
得 9p = 18
得 p = 2
帶回
x = 3,
y = 4,
z =11,
(1)x=?
x = 3
(2)( 2x+y):(z+5)=?
= (2*3+4) : (11+5)
= 10 : 16
= 5 : 8
2011-05-06 7:40 am
(x+1)：(y+2)：(z－3)=2：3：4，且x+y+z=18則：
(1)x=?
(x+1)：(y+2)：(z－3)=2：3：4
(x+1)/2=(y+2)/3=(z－3)/4
(x+1)/2=(y+2)/3=(z－3)/4=[(x+1)+(y+2)+(z-3)]/(2+3+4)=18/9=2
x+1=4，x=3
y=2=6，y=4
z－3=8，z=11
(2)(2x+y)：(z+5)=?
(2x+y):(z+5)=(6+4)：(11+5)=5：8
(1)x=?
(x+1)：(y+2)：(z－3)=2：3：4
(x+1)/2=(y+2)/3=(z－3)/4
(x+1)/2=(y+2)/3=(z－3)/4=[(x+1)+(y+2)+(z-3)]/(2+3+4)=18/9=2
x+1=4，x=3
y=2=6，y=4
z－3=8，z=11
(2)(2x+y)：(z+5)=?
(2x+y):(z+5)=(6+4)：(11+5)=5：8
2011-05-06 7:20 am
(x+1):(y+2):z-3=2:3:4,且x+y+z=18則:
(1)x=?令x+1=2t
=>x=2t-1
y+2=3t
=>y=3t-2
z-3=4t
=>z=4t+3x+y+z=2t-1+3t-2+4t+3
=>18=9t
=>t=2x
=2t-1
=3#(2)(2x+y):(z+5)=?2x+y:z+5
=10:16
=5:8#有誤請提醒
Thanks~
(1)x=?令x+1=2t
=>x=2t-1
y+2=3t
=>y=3t-2
z-3=4t
=>z=4t+3x+y+z=2t-1+3t-2+4t+3
=>18=9t
=>t=2x
=2t-1
=3#(2)(2x+y):(z+5)=?2x+y:z+5
=10:16
=5:8#有誤請提醒
Thanks~
收錄日期: 2021-04-30 15:46:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110505000015KK08911