moon_4
幼苗
共回答了15个问题采纳率:100% 举报
一大题:1/(ab+a+1)+1/(bc+b+1)+1/(ca+c+1)
=abc/(ab+a+abc)+1/(bc+b+1)+1/(ca+c+1)······第一项的分子分母的1用abc代替;
=bc/(b+1+bc)+1/(bc+b+1)+1/(ca+c+1)
=(bc+1)/(bc+b+1)+1/(ca+c+1)
=(bc+abc)/(bc+b+abc)+1/(ca+c+1))······第一项的分子分母的1用abc代替;
=(c+ca)/(c+1+ca)+1/(ca+c+1)
=(ca+c+1)/(ca+c+1)
=1
二大题
因为 ab/(a+b)=1/3 ,bc/(b+c)=1/4 ,ca/(c+a)=1/5
所以:
(a+b)/ab = 3
(b+c)/bc = 4
(a+c)/ac = 5
即:
1/a + 1/b = 3
1/b + 1/c = 4
1/a + 1/c = 5
三式相加,得:
2(1/a + 1/b + 1/c) = 12
所以:1/a + 1/b + 1/c = 6
先求“abc/(ab+bc+ca)”的倒数:
(ab+bc+ca)/abc
= 1/a + 1/b + 1/c = 6
所以:
abc/(ab+bc+ca) = 1/6
1年前
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