根号a-1加根号b-2等于0,求1/ab+1/(a+)(b+1)+1/(a+2)(b+2)+1/(a+3)(b+3)……1/(a+2008)(b+2008)
人气:319 ℃ 时间:2020-02-05 13:19:37
解答
根号求解是非负数~两个非负数相加要等于0,只有两个都为0所以根号a-1=0根号b-2=0即a=1,b=2然后代进表达式,就可以求解了!表达式分解下=[1/(b-a)]*[(1/a)-(1/b)]+[1/(b-a)]*[(1/(a+1))-(1/(b+1))]+...=(1-1/2) + 1/2 *(...
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