已知A、B、C为实数且满足ab/(a+b)=1/3,bc/(b+c)=1/4,ac/(a+c)=1/5,求abc/(ab+bc+ac)的值
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![已知A、B、C为实数且满足ab/(a+b)=1/3,bc/(b+c)=1/4,ac/(a+c)=1/5,求abc/(ab+bc+ac)的值](/uploads/image/z/4528796-68-6.jpg?t=%E5%B7%B2%E7%9F%A5A%E3%80%81B%E3%80%81C%E4%B8%BA%E5%AE%9E%E6%95%B0%E4%B8%94%E6%BB%A1%E8%B6%B3ab%2F%28a%2Bb%29%3D1%2F3%2Cbc%2F%28b%2Bc%29%3D1%2F4%2Cac%2F%28a%2Bc%29%3D1%2F5%2C%E6%B1%82abc%2F%28ab%2Bbc%2Bac%29%E7%9A%84%E5%80%BC)
已知A、B、C为实数且满足ab/(a+b)=1/3,bc/(b+c)=1/4,ac/(a+c)=1/5,求abc/(ab+bc+ac)的值
已知A、B、C为实数且满足ab/(a+b)=1/3,bc/(b+c)=1/4,ac/(a+c)=1/5,求abc/(ab+bc+ac)的值
已知A、B、C为实数且满足ab/(a+b)=1/3,bc/(b+c)=1/4,ac/(a+c)=1/5,求abc/(ab+bc+ac)的值
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 + ...
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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
1/a + 1/b + 1/c
=(ab+bc+ca)/abc
=(ab+bc+ca)/abc
= 6
取倒数:abc/(ab+bc+ca) = 1/6
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已知的分别倒数后1/a+1/b=3 1/b+1/c=4 1/a+1/c=5
三式相加除以2得:1/a+1/b+1/c=6
abc/(ab+bc+ac)=1/(1/c+1/b+1/a)=1/6
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