|ab-2|+(1-b)^2=0 则|ab-2|=(1-b)^2=0 则ab=2,b=1 则a=2,b=1 则1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2008(b+2008) =1/(1*2)+1/(2*3)+...+1/(2009*2010) =1/1-1/2+1/2-1/3+...+1/2009-1/2010=1-1/2010 =2009/2010...
如果有理数a,b满足|ab-2|+|1-b|=0试求:
4个回答
相关问题
-
如果有理数a,b满足|ab-2|+(1-b)2=0;试求[1/ab]+[1(a+1)(b+1)
-
如果有理数a、b满足|ab-2|+(1-b)^2 =0,试求1/ab+1/(a+1)*(b+1)+
-
如果有理数a、b满足|ab-2|+|1-b|=0,试求1 /ab+1/a+1)(b+1)+1/(a+2)(b+2)+..
-
如果有理数a,b满足|ab-2|+|1-b|=0,试求1 /ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.
-
如果有理数a,b满足|ab-2|+(1-b) 2 =0,试求 1 ab + 1 (a+1)(b+1) + 1 (a+2)
-
如果有理数a,b满足|ab-2|+|1-b|=0.试求1/ab+1/(a+1)(b+1)+1(a+2)(b+2)+…+1
-
如果有理数a和b满足|ab-2|+|1-b|=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+..
-
如果有理数a,b满足|ab-2|+(1-b)×(1-b)=0,试求1/ab + 1/(a+1)(b+1) + 1/(a+
-
如果有理数a,b满足丨ab-2丨+(1-b)^2=0 试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)
-
如果有理数a,b满足|ab-2|+(1-6)的平方=0试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+