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)+1/(A+2)(B+2)…+1/(A+2001)(B+2004)

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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)+1/(A+2)(B+2)…+1/(A+2001)(B+2004)

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)+1/(A+2)(B+2)…+1/(A+2001)(B+2004)
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)+1/(A+2)(B+2)…+1/(A+2001)(B+2004)

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)+1/(A+2)(B+2)…+1/(A+2001)(B+2004)
由题可知AB-2=0,1-B=0 所以AB=2 B=1,可以知道A=2
原式=1-1/2+1/2-1/3.+1/2003-1/2005
=1-1/2005
=2004/2005

偶才小学五年级!

由题可知AB-2=0,1-B=0 所以AB=2 B=1, 可以知道A=2
原式等于1/2+1/6+1/12......
1/2=1/1-1/2,1/6+1/2-1/3,1/12=1/3-1/4......
所以原式等于1-1/2+1/2-1/3+1/3-1/4......
最后可以化简
希望我没做错!!!

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