求一元一次方程去分母题40道

1个回答

  • 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+12(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 3x+7=32-2x

    3x+5(138-x)=540

    3x-7(x-1)=3-2(x+3)

    18x+3x-3=18-2(2x-1)

    3(20-y)=6y-4(y-11)

    -(x/4-1)=5

    3[4(5y-1)-8]=6

    10/3 (x/5+3/7)=9x/2

    5/3(x+0.5)+2=3x-6

    5x+2(2x/3+2)=2/3(x-6)+2

    (2x-7)/2-(6x-5)/3=2x+3

    (3x+2)/5-(x-6)=x/3

    6x-(x/3+2)=2(x/5+5/2)-3

    3(x/11-2)-5=2+3x/3

    10/3(2x-6)=3/5

    x/2-(x/3-2)=3

    2/3(x+3)-3=5x/3

    5/3(2x-5/3)=2x/5-8/9

    25(x/3-x/2+2/5)-2=3/5(x-2/7)+4/9

    5x*56+(-3^3-x)]/9=5

    89x/3-5^2-(8-5x)/5=541

    x+7-(-36+8^2)/2=8+7^4/3

    a-7-98+7a=3.2*5a

    89/2+35/6x=3*9+2^3/5+7x

    3X+189/3=521/2

    4Y+119*^3=22/11

    7(2x-1)-3(4x-1)/9=[4(3x+2)-1]/9

    [(5y+1)+ (1-y)]/2= [(9y+1)+ (1-3y)]/3

    [-6(-7^4*8)-4]/5=(x+2)/6

    2/3*8*1/4x=89/2

    20%/5+(1-20%)(320-x)/9=320×40%/3

    2(x-2)/6+2/9=(x+1)/2

    2(x-2)/2-3(4x-1)/3=9(1-x)/2

    11x/2+(64-2x)/6=(100-9x)/8

    15-(8-5x)/2=7x/3+(4-3x)/4

    3(x-7)/4-2[9-4(2-x)]/9=22/3

    3/2[2/3(1/4x-1)-2]-x/9=2/5

    2x+7^2/2=157/5