一元一次方

2009-12-20 7:08 pm
解下列各方程
1. (5x+9)/2=4x
2. (16-3x)/5= -3/x
3. [5 (x+1)] / 3=(5x+10) / 8
4. (x-2)/5-(x+2)/2=4
更新1:

更正: 第2.條打錯 應該係:(16-3x)/5= -(x/3)

回答 (3)

2009-12-20 7:15 pm
✔ 最佳答案
1. (5x+9)/2=4x
5x+9=8x
3x=9
x=3

3. [5 (x+1)] / 3=(5x+10) / 8
40(x+1)=3(5x+10)
40x+40=15x+30
25x=-10
x=-2/5

4. (x-2)/5-(x+2)/2=4
2(x-2)-5(x+2)=40
2x-4-5x-10=40
-3x=54
x=-18

2009-12-20 11:15:57 補充:
第2題不是一元一次,你有沒有打錯?!

2009-12-21 22:51:51 補充:
(16-3x)/5= -(x/3)
3(16-3x)=-5x
48-9x=-5x
48=4x
x=12
2009-12-29 2:18 am
1. (5x+9)/2=4x
(5x+9)/2 (2)=4x(2)
5x+9=8x
5x-8x=-9
-3x=-9
x=-9/-3
x=3

2. (16-3x)/5= -(x/3)
(16-3x)/5 (5)= -(x/3)(5)
16-3x=-(5x/3)
3(16-3x)=-(5x/3)(3)
16(3)+(-3x)(3)=-5x
48-9x=-5x
-9x+5x=-48
-4x=-48
x=-48/-4
x=12

3. [5 (x+1)] / 3=(5x+10) / 8
[5x+5(1)]/3=(5x+10)/8
(5x+5)/3=(5x+10)/8
(5x+5)/3 (24)=(5x+10)/8 (24)
8(5x+5)=3(5x+10)
40x+40=15x+30
40x-15x=30-40
25x=-10
x=-10/25
x=-0.4

4. (x-2)/5-(x+2)/2=4
[(x-2)/5-(x+2)/2](10)=4(10)
(x-2)/5 (10)+[-(x+2)/2](10)=40
2(x-2)-5(x+2)=40
2x-4-5x-10=40
-3x-14=40
-3x=40+14
-3x=54
x=54/-3
x=-18



















2009-12-23 3:28 am
1. (5x+9)/2=4x
5x+9=8x
9=8x-5x
3=3x
3x=3
x=3/3
x=1

2. (16-3x) / 5= - (x / 3)
5[(16-3x) / 5]=5[-(x / 3)]
16-3x= - 5(x / 3)
3(16-3x)= 3[- (5x / 3)]
48-9x= - 5x
48-9x+9x= - 5x+9x
48=4x
4x=48
x=12

3. [5 (x+1)] / 3=(5x+10) / 8
(5x+5) / 3 = (5x+10) / 8
24[(5x+5)/3]= 24[(5x+10) / 8]
8(5x+5)=3(5x+10)
40x+40=15x+30
40x-15x=30-40
25x= - 10
x= - 5/2
x= - 2.5

4. (x-2)/5-(x+2)/2=4
10[(x-2)/5] - 10[(x+2)/2] = 4*10
2(x-2) - 5(x+2) = 40
2x-4-5x-10 = 40
-3x-14 = 40
-3x = 54
x=54/(-3)
x=-18
參考: Myself


收錄日期: 2021-04-23 20:42:24
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