初一数学:计算:(1.)(2分之1-3+2又6分之5-1又12分之7)÷(-36分之1)

2个回答

  • .(2分之1-3+2又6分之5-1又12分之7)÷(-36分之1)

    =(1/2-3+2+5/6-1-7/12)×(-36)

    =(1/2-2+5/6-7/12)×(-36)

    =-18+72-30+21

    =45

    -3³-[-5-0.2÷5分之4×(-2)²]

    =-27+5+1/5×5/4×4

    =-27+5+1

    =-21

    12分之10-13x=1-3分之2x-1

    10-13x=12-4(2x-1)

    10-13x=12-8x+4

    -13x+8x=16-10

    -5x=6

    x=-6/5

    0.01分之0.1x-0.02-0.5分之x+1=3

    10x-2-5分之(10x+10)=3

    50x-10-10x-10=15

    40x=45

    x=9/8

    5.已知(x+1)²+|y-2分之1|=0,求2(xy²+x²y)-[2xy²-3(1-x²y)]-2的值

    (x+1)²+|y-2分之1|=0

    x+1=0 y-1/2=0

    x=-1 y=1/2

    2(xy²+x²y)-[2xy²-3(1-x²y)]-2

    =2xy²+2x²y-2xy²+3-3x²y-2

    =-x²y+1

    =-(-1)²*1/2+1

    =-1/2+1

    =1/2