|ab-2|与|b-1|互为相反数,求下列代数式的值 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2).+1/(a+2009)(b+2009)
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|ab-2|与|b-1|互为相反数,求下列代数式的值 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2).+1/(a+2009)(b+2009)
|ab-2|与|b-1|互为相反数,求下列代数式的值 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2).
+1/(a+2009)(b+2009)
|ab-2|与|b-1|互为相反数,求下列代数式的值 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2).+1/(a+2009)(b+2009)
由题:b=1,ab=2,a=2
原式=1/2+1/(2*3)+1/(3*4)+…1/(2010*2011)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+…(1/2010-1/2011)
=1-1/2011
=2010/2011
∵|ab-2|与|b-1|互为相反数
|ab-2|≥0,|b-1|≥0
∴ab-2=0;
b-1=0;
b=1;a=2;
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)......+1/(a+2009)(b+2009)
=1/1*2+1/(2*3)+1/(3*4)+...+1/(2010*2011)
=1-1/2+1/2-1/3+1/3-1/4+....+1/2010-1/2011
=1-1/2011
=2010/2011
由于绝对值必为非负数,且两个绝对值互为相反数,所以ab-2=b-1=0,所以a=2,b=1,接下来只需要裂项一次即可,所以原式=2010/2011
解
因为|ab-2|≥0 |b-1|≥0
又|ab-2|与|b-1|互为相反数
所以ab-2=0 b-1=0
解得a=2 b=1 ab=2
所以
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)...... +1/(a+2009)(b+2009)
=1/1*2+1/2*3+1/3*4+......+1/2010*2011
=(1-1/2)+(12-1/3)+(1/3-1/4)+.....+(1/2010-1/2011)
=1-1/2011
=2010/2011
|ab-2|与|b-1|互为相反数,可得:
|ab-2|+|b-1|=0 且有:|ab-2|≥0,|b-1|≥0
所以可得:
ab-2=0
b-1=0 解得:a=2,b=1
所以有:
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)......+1/(a+2009)(b+2009)
=1/1x2+1/2x3+1/3x4+····+...
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|ab-2|与|b-1|互为相反数,可得:
|ab-2|+|b-1|=0 且有:|ab-2|≥0,|b-1|≥0
所以可得:
ab-2=0
b-1=0 解得:a=2,b=1
所以有:
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)......+1/(a+2009)(b+2009)
=1/1x2+1/2x3+1/3x4+····+1/2010x2011
=1-1/2+1/2-1/3+1/3-1/4+···+1/2010-1/2011
=1-1/2011
=2010/2011
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