急急2題數學題.解下列各方程(10分)

2007-12-10 2:14 am
4x-[7-2(x-1)]=x+11

3[3(4-x)+1]=7[2(3-2x)+5]

回答 (6)

2007-12-10 2:22 am
✔ 最佳答案
4x-[7-2(x-1)]=x+11
4x-(7-2x+2)=x+11
4x-5+2x= x+11
6x-5=x+11
5x=16
x = 16/5
= 3/1/5
3[3(4-x)+1]=7[2(3-2x)+5]
3(12-3x+1) = 7(6-4x+5)
3(13-3x) = 7(11-4x)
39-9x = 77-28x
17x = 38
x = 38/17
= 2/4/17
2007-12-15 10:04 pm
4x-[7-2(x-1)]=x+11
4x-(5x-5)=x+11
3x=5x-5+11
3x=5x+6
2x=-6
x=-3

3[3(4-x)+1]=7[2(3-2x)+5]
3(12-3x+1)=7(6-4x+5)
3(13-3x)=7(11-4x)
39-9x=77-28x
19x=38
x=2
參考: me
2007-12-10 2:31 am
你要先化簡括號d數
4x-[7-2(x-1)]=x+11
4x-[7-2x+2]=x+11
4x-7+2x-2 =x+11
4x+2x-x =11+7+2
5x =20
x =4



3[3(4-x)+1]=7[2(3-2x)+5]
3[12-3x+1]=7[6-4x+5]
36-9x+3=42-28x+35
-9x+28x=42+35-36-3
19x=38
x=2
參考: 自己
2007-12-10 2:29 am
1. 4x-[7-2(x-1)]=x+11
4X-(7-2X+2)=x+11
4X-7+2x-2 =X+11
6x-9 =x+11
-9-11 =X-6x
-20=-5X
x=4

2. 3[3(4-x)+1]=7[2(3-2x)+5]
3(12-3x+1)=7(6-4X+5)
39-9x =77-28X
39-77=-28X+9x
-38=-19x
x=2
2007-12-10 2:27 am
4x-[7-2(x-1)]=x+11
4x-(7-2x+2)=x+11
4x-9+2x=x+11
6x-9=x+11
5x=20
x=4

3[3(4-x)+1]=7[2(3-2x)+5]
3(12-3x+1)=7(6-4x+5)
3(13-3x)=7(11-4x)
39-9x=77-28x
19x=38
x=2
參考: 自己!
2007-12-10 2:24 am
1.
4x-[7-2(x-1)]=x+11
4x-(7-2x+2)=x+11
4x-(9-2x)=x+11
4x-9+2x=x+11
6x-9=x+11
(6-1)x=11+9
5x=20
x=4
2.

3[3(4-x)+1]=7[2(3-2x)+5]
3(12-3x+1)=7(6-4x+5)
3(13-3x)=7(11-4x)
39-9x=77-28x
(28-9)x=77-39
19x=38
x=2

2007-12-09 18:27:21 補充:
樓上計錯數............4x-[7-2(x-1)]=x 114x-(7-2x 2)=x 114x-5 2x= x 11[呢度計錯]應該係{4x-(9-2x)}=4x-[9] 2x唔係-56x-5=x 115x=16x = 16/5 = 3/1/5

2007-12-09 18:27:26 補充:
3[3(4-x) 1]=7[2(3-2x) 5] 3(12-3x 1) = 7(6-4x 5)3(13-3x) = 7(11-4x)39-9x = 77-28x 17x = 38 [呢度計錯...]{28-9}x=19x唔係17x x = 38/17   = 2/4/17


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