If the rational numbers a and B satisfy | ab-2 | + (1-B) & # 178; = 0; Try to find 1 / AB + (a + 1) (B + 1) 1 / AB + +(a + 2010) (B + 2010) 1 +?

If the rational numbers a and B satisfy | ab-2 | + (1-B) & # 178; = 0; Try to find 1 / AB + (a + 1) (B + 1) 1 / AB + +(a + 2010) (B + 2010) 1 +?

|ab-2|+（1-b）²=0
So ab-2 = 0
1-b=0
So B = 1
a=2/b=2
So the original formula = 1 / 1 × 2 + 1 / 2 × 3 + +1/2011×2012
=1-1/2+1/2-1/3+…… +1/2011-1/2012
=1-1/2012
=2011/2012

Given that rational numbers a and B satisfy AB & # 178; < 0, a + b > 0, and ︱ a = 2, ︱ B = 3, find the value of ︱ a - &# 8531; ︱ + (B-1) &# 178

Because B & # 178; > 0
So A0
So only b > 0
So a = - 2, B = 3
So the original formula = | - 2-1 / 3 | + (3-1) & # - 178;
=7/3+4
=6 and 1 / 3

If a and B are rational numbers, and | A-1 | + (ab-2) & # 178; = 0, find 1 of AB + (a + 1) (B + 1) 1 of AB + (a + 2) (B + 2) + +(a + 1009) (B + 2009)

Because, a and B are rational numbers, and | A-1 | + (ab-2) & # 178; = 0
So, A-1 = 0 and ab-2 = 0,
The solution is a = 1, B = 2, which is substituted into the formula
The formula = 1 / (1x2) + 1 / (2x3) + 1 / (3x4) +. + 1 / (2010x2011)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.+(1/2010-1/2011)
=1+(-1/2+1/2)+(-1/3+1/3)+(-1/4+1/4)+.+(-1/2010+1/2010)-1/2011
=1-1/2011
=2010/2011
It should be 2009 + 100A