|ab-2|+(1-b)的平方=0
∴ab-2=0
1-b=0
∴a=2
b=1
求ab份之1+(a+1)*(b+1)份之1+(a+2)*(b+2)份之1
=1/1×2+1/2×3+1/3×4
=1-1/2+1/2-1/3+1/3-1/4
=1-1/4
=3/4
|ab-2|+(1-b)的平方=0
∴ab-2=0
1-b=0
∴a=2
b=1
求ab份之1+(a+1)*(b+1)份之1+(a+2)*(b+2)份之1
=1/1×2+1/2×3+1/3×4
=1-1/2+1/2-1/3+1/3-1/4
=1-1/4
=3/4