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