有理数A,B满足|AB-2|+(1-B)^2=0,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+(A+2007)(,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+1/(A+2007)(B+2007)=?
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![有理数A,B满足|AB-2|+(1-B)^2=0,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+(A+2007)(,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+1/(A+2007)(B+2007)=?](/uploads/image/z/1812253-13-3.jpg?t=%E6%9C%89%E7%90%86%E6%95%B0A%2CB%E6%BB%A1%E8%B6%B3%7CAB-2%7C%2B%281-B%29%5E2%3D0%2C%E6%B1%821%2FAB%2B1%2F%28A%2B1%29%28B%2B1%29%2B1%2F%28A%2B2%29%28B%2B2%29%2B...%2B%28A%2B2007%29%28%2C%E6%B1%821%2FAB%2B1%2F%28A%2B1%29%28B%2B1%29%2B1%2F%28A%2B2%29%28B%2B2%29%2B...%2B1%2F%28A%2B2007%29%28B%2B2007%29%3D%3F)
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有理数A,B满足|AB-2|+(1-B)^2=0,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+(A+2007)(,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+1/(A+2007)(B+2007)=?
有理数A,B满足|AB-2|+(1-B)^2=0,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+(A+2007)(
,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+1/(A+2007)(B+2007)=?
有理数A,B满足|AB-2|+(1-B)^2=0,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+(A+2007)(,求1/AB+1/(A+1)(B+1)+1/(A+2)(B+2)+...+1/(A+2007)(B+2007)=?
因为|AB-2|+(1-B)^2=0;
而 |AB-2|>=0 且 (1-B)^2>=0;
则要满足条件,只能,
|AB-2|=0 且 (1-B)^2=0;
即 AB=2 且 B=1 →A=2;B=1
所以原式可化为
1/2+1/(2*3)+1/(3*4)+1/(4*5)+...+1/(2008*2009)
=1/2+(3-2)/(2*3)+(4-3)/(3*4)+(5-4)/(4*5)+...+(2009-2008)/(2008*2009)
=1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/2008-1/2009
=1/2+1/2-1/2009
=1-1/2009
=2008/2009
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