如果有理数a,b满足|ab-2|+(1-b)的2次方=0 试求1/ab+1/(a+a)(b+1)+1(a+2)(b+2)
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如果有理数a,b满足|ab-2|+(1-b)的2次方=0 试求1/ab+1/(a+a)(b+1)+1(a+2)(b+2)+...+1/(a+2004)(b+2004)的
“/” 为分数线
不是1/ab+1/(a+a)(b+1)+1(a+2)(b+2)+...+1/(a+2004)(b+2004)
是1/ab+1/(a+1)(b+1)+1(a+2)(b+2)+...+1/(a+2004)(b+2004)
“/” 为分数线
不是1/ab+1/(a+a)(b+1)+1(a+2)(b+2)+...+1/(a+2004)(b+2004)
是1/ab+1/(a+1)(b+1)+1(a+2)(b+2)+...+1/(a+2004)(b+2004)
绝对值和平方都大于等于0
先放假为0则都等于0
所以ab-2=0,1-b=0
b=1,a=2/b=2
所以原式=1/1*2+1/2*3+1/3*4+……+1/2005*2006
=1-1/2+1/2-1/3+1/3-1/4+……+1/2005-1/2006
=1-1/2006
=2005/2006
先放假为0则都等于0
所以ab-2=0,1-b=0
b=1,a=2/b=2
所以原式=1/1*2+1/2*3+1/3*4+……+1/2005*2006
=1-1/2+1/2-1/3+1/3-1/4+……+1/2005-1/2006
=1-1/2006
=2005/2006
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