【求一元一次方程20道,全部含有分母的过程】
1人问答
问题描述：
求一元一次方程
20道,全部含有分母的过程
谭焕云回答：
7(2x-1)-3(4x-1)=4(3x+2)-1;
(5y+1)+(1-y)=(9y+1)+(1-3y);
20%+(1-20%)(320-x)=320×40%
2(x-2)+2=x+1
2(x-2)-3(4x-1)=9(1-x)
x/3-5=(5-x)/2
2(x+1)/3=5(x+1)/6-1
(1/5)x+1=(2x+1)/4
(5-2)/2-(4+x)/3=1
x/3-1=(1-x)/2
(x-2)/2-(3x-2)/4=-1
11x+64-2x=100-9x
15-(8-5x)=7x+(4-3x)
3(x-7)-2[9-4(2-x)]=22
3/2[2/3(1/4x-1)-2]-x=2
2(x-2)-3(4x-1)=9(1-x)
11x+64-2x=100-9x
15-(8-5x)=7x+(4-3x)
3(x-7)-2[9-4(2-x)]=22
3/2[2/3(1/4x-1)-2]-x=2
2(x-2)+2=x+1
1.7(2x-1)-3(4x-1)=4(3x+2)-1
2.(5y+1)+(1-y)=(9y+1)+(1-3y)
3.[(-2)-4]=x+2
4.20%+(1-20%)(320-x)=320×40%
5.2(x-2)+2=x+1
6.2(x-2)-3(4x-1)=9(1-x)
7.11x+64-2x=100-9x
8.15-(8-5x)=7x+(4-3x)
9.3(x-7)-2[9-4(2-x)]=22
10.3/2[2/3(1/4x-1)-2]-x=2
11.5x+1-2x=3x-2
12.3y-4=2y+1
13.87X*13=5
14.7Z/93=41
15.15X+863-65X=54
16.58Y*55=27489
17.2（x+2）+4=9
18.2(x+4)=10
19.3(x-5)=18
20.4x+8=2(x-1)
21.3(x+3)=9+x
22.6(x/2+1)=12
23.9(x+6)=63
24.2+x=2(x-1/2)
25.8x+3(1-x)=-2
26.7+x-2(x-1)=1
27.x/3-5=(5-x)/2
28.2(x+1)/3=5(x+1)/6-1
29.(1/5)x+1=(2x+1)/4
30.(5-2)/2-(4+x)/3=1
15x-8(5x+1.5)=18*1.25+x
3X+189=521
4Y+119=22
3X*189=5
8Z/6=458
3X+77=59
4Y-6985=81
87X*13=5
7Z/93=41
15X+863-65X=54
58Y*55=27489
1.2(x-2)-3(4x-1)=9(1-x)
2.11x+64-2x=100-9x
3.15-(8-5x)=7x+(4-3x)
4.3(x-7)-2[9-4(2-x)]=22
5.3/2[2/3(1/4x-1)-2]-x=2
6.2(x-2)+2=x+1
7.0.4(x-0.2)+1.5=0.7x-0.38
8.30x-10(10-x)=100
9.4(x+2)=5(x-2)
10.120-4(x+5)=25
11.15x+863-65x=54
12.12.3(x-2)+1=x-(2x-1)
13.11x+64-2x=100-9x
14.14.59+x-25.31=0
15.x-48.32+78.51=80
16.820-16x=45.5×8
17.(x-6)×7=2x
18.3x+x=18
19.0.8+3.2=7.2
20.12.5-3x=6.5
21.1.2(x-0.64)=0.54
22.x+12.5=3.5x
23.8x-22.8=1.2
24.150x+10=60
25.260x-30=20
26.33^20x+50=110
27.42x=5x-3
28.590=10+x
29.690+20x=30
30.7691+3x=700
12x-10.3x=15
20.52x-(1-0.52)x=80
3x/2+3x/2=7
43x+7=32-2x
53x+5(138-x)=540
63x-7(x-1)=3-2(x+3)
718x+3x-3=18-2(2x-1)
83(20-y)=6y-4(y-11)
9-(x/4-1)=5
103[4(5y-1)-8]=6
没有过程了,没有人会写过程的就多给点题和答按了,给点分吧
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