.(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 .(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