已知|a-1|+(ab-2)²=0,求1/ab + 1/(a+1)(b+1) + 1/(a+2)(b+2) + 1/(a+3)(b+3)的值.
问题描述：
已知|a-1|+(ab-2)²=0,求1/ab + 1/(a+1)(b+1) + 1/(a+2)(b+2) + 1/(a+3)(b+3)的值.
答
a-1=0, ab-2=0
a=1, b=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+1/(a+3)(b+3)
=1/1*2+1/2*3+1/3*4+1/4*5
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5
=1-1/5
=4/5
答
已知|a-1|+(ab-2)²=0
则，a-1=0 ab-2=0
a=1 b=2
原式=1/2+1/6+1/12+1/20
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5
=1-1/5
=4/5
答
a-1|+(ab-2)²=0,可得a=1,b=2.
代入：1/(1×2）-1/(2×3）-1/(3×4）-1/(4×5）=1-1/2+1/2-1/3+1/3-1/4-1/4+1/4-1/5=4/5