Question

已知|ab=2|+(a+1)^2=0,求下列式子的值：1/（a-1)(b+1)+1\(a-2)(b+2)+……+1/(a

Answer 1

已知|ab+2|+(a+1)^2=0

因为|ab+2|≥0,(a+1)^2≥0

所以,当它们之和为零时,必然是两者同时为零

所以,ab+2=0,a+1=0

所以,a=-1,b=2

所以原式=1/[(-2)*3]+1/[(-3)*4]+……+1/[(-101)*102]

=-[1/(2*3)+1/(3*4)+……+1/(101*102)]

=-[(1/2)-(1/3)+(1/3)-(1/4)+……+(1/101)-(1/102)]

=-[(1/2)-(1/102)]

=-25/51