1樓：科學普及交流
a/(ab+a+1) +b/(bc+b+1) +c/(ca+c+1)=a/(ab+a+abc)+b/(bc+b+1)+c/(ca+c+abc)
=1/(b+1+bc)+b/(bc+b+1)+1/(a+ab+1)=1/(b+1+bc)+b/(bc+b+1)+abc/(a+ab+abc)
=1/(b+1+bc)+b/(bc+b+1)+bc/(bc+b+1)=(bc+b+1)/(bc+b+1)=1
2樓：
ab+a+1=ab+a+abc=a(b+1+bc)=a(b+abc+bc)=ab(1+ac+c),
所以a/(ab+a+1)=1/b(1+ac+c)=ac/(1+ac+c)
同理，b/(bc+b+1)=1/(1+ac+c)所以a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)=(ac+1+c)/(ac+c+1)=1
3樓：濤濤老師
回答親親您好，非常高興能回答您的問題:
1=abc
a/(ab+a+1) + b/(bc+b+1) + c/(ac+c+1)
= a/(ab+a+abc) + b/(bc+b+1) + c/(ac+c+1)
= 1/(b+1+bc) + b/(bc+b+1) + c/(ac+c+1)
= 1/(bc+b+1) + b/(bc+b+1) + c/(ac+c+1)
= (1+b)/(bc+b+1) + c/(ac+c+1)
= (abc+b)/(bc+b+abc) + c/(ac+c+1)
= (ac+1)/(c+1+ac) + c/(ac+c+1)
= (ac+1)/(ac+c+1) + c/(ac+c+1)
= (ac+1+c)/(ac+c+1)
= 1更多10條
若abc=1,求ab+a+1分之a+bc+b+1分之b+a+c+1分之c的值
4樓：匿名使用者
最後一項寫錯了，應該是(ca+c+1)分之c。
a/(ab+a+1) +b/(bc+b+1) +c/(ca+c+1)=a/(ab+a+abc)+b/(bc+b+1)+c/(ca+c+abc)
=1/(b+1+bc)+b/(bc+b+1)+1/(a+ab+1)=1/(b+1+bc)+b/(bc+b+1)+abc/(a+ab+abc)
=1/(b+1+bc)+b/(bc+b+1)+bc/(bc+b+1)=(bc+b+1)/(bc+b+1)=1
5樓：匿名使用者
解：a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)=a/(ab+a+abc)+b/(bc+b+1)+c/(ac+c+1)
=1/(b+1+bc)+b/(bc+b+1)+c/(ac+c+1)=(1+b)/(bc+b+1)+c/(ac+c+1)=(abc+b)/(bc+b+abc)+c/(ac+c+1)=(ac+1)/(c+1+ac)+c/(ac+c+1)=(ac+c+1)/(ac+c+1)=1
6樓：濤濤老師
回答親親您好，非常高興能回答您的問題:
1=abc
a/(ab+a+1) + b/(bc+b+1) + c/(ac+c+1)
= a/(ab+a+abc) + b/(bc+b+1) + c/(ac+c+1)
= 1/(b+1+bc) + b/(bc+b+1) + c/(ac+c+1)
= 1/(bc+b+1) + b/(bc+b+1) + c/(ac+c+1)
= (1+b)/(bc+b+1) + c/(ac+c+1)
= (abc+b)/(bc+b+abc) + c/(ac+c+1)
= (ac+1)/(c+1+ac) + c/(ac+c+1)
= (ac+1)/(ac+c+1) + c/(ac+c+1)
= (ac+1+c)/(ac+c+1)
= 1更多10條
若abc=1則ab+a+1分之a加bc+b+1分之b加ac+c+1之c等於多少
7樓：天堂蜘蛛
解：因為abc=1
所以原式=(abc/ab^2c+abc+bc)+(abc/abc^2+abc+ac)+(abc/a^2bc+abc+ab)
=(1/bc+b+1)+(1/ac+c+1)+(1/ab+a+1)=((1/bc+b+1)+(b/abc+bc+b）+(bc/ab^2c+abc+bc)
=(1/bc+b+1)+(b/bc+b+1)+(bc/bc+b*abc+1)
=(bc+b+1)(bc+b+1)
=1所以所求代數式的值是1
8樓：匿名使用者
a/(ab+a+1) + b/(bc+b+1) + c/(ac+c+1)
= a/(ab+a+abc) + b/(bc+b+1) + bc/(abc+bc+b)
= 1/(b+1+bc) + b/(bc+b+1) + bc/(1+bc+b)
= (1+b+bc)/(b+1+bc)= 1
已知abc=1，求ab+a+1分之a加上bc+b+1分之b加上ac+c+1分之c的值
9樓：匿名使用者
abc=1
所以b=1/ac
ab=1/c
bc=1/a
所以原式=a/(1/c+a+1)+(1/ac)/(1/a+1/ac+1)+c/(ac+c+1)
第乙個式子上下同乘c
第二個式子上下同乘ac
所以=ac/(ac+c+1)+1/(ac+c+1)+c/(ac+c+1)
=(ac+c+1)/(ac+c+1)=1
10樓：
注意是abc=1的帶換
a/(ab a 1) b/(bc b 1) c/(ac c 1)=1/(b 1 bc) b/(bc b 1) c/(ac c 1)=(1 b)/(b 1 bc) c/(ac c 1)=(1 ac c)/(1 ac c)=1
11樓：濤濤老師
回答親親您好，非常高興能回答您的問題:
1=abc
a/(ab+a+1) + b/(bc+b+1) + c/(ac+c+1)
= a/(ab+a+abc) + b/(bc+b+1) + c/(ac+c+1)
= 1/(b+1+bc) + b/(bc+b+1) + c/(ac+c+1)
= 1/(bc+b+1) + b/(bc+b+1) + c/(ac+c+1)
= (1+b)/(bc+b+1) + c/(ac+c+1)
= (abc+b)/(bc+b+abc) + c/(ac+c+1)
= (ac+1)/(c+1+ac) + c/(ac+c+1)
= (ac+1)/(ac+c+1) + c/(ac+c+1)
= (ac+1+c)/(ac+c+1)
= 1更多10條
abc=1，求,ab+a+1分之a+bc+b+1分之b+ac+c+1分之c的值
12樓：買昭懿
1=abc
a/(ab+a+1) + b/(bc+b+1) + c/(ac+c+1)
= a/(ab+a+abc) + b/(bc+b+1) + c/(ac+c+1)
= 1/(b+1+bc) + b/(bc+b+1) + c/(ac+c+1)
= 1/(bc+b+1) + b/(bc+b+1) + c/(ac+c+1)
= (1+b)/(bc+b+1) + c/(ac+c+1)
= (abc+b)/(bc+b+abc) + c/(ac+c+1)
= (ac+1)/(c+1+ac) + c/(ac+c+1)
= (ac+1)/(ac+c+1) + c/(ac+c+1)
= (ac+1+c)/(ac+c+1)= 1
13樓：濤濤老師
回答親親您好，非常高興能回答您的問題:
1=abc
a/(ab+a+1) + b/(bc+b+1) + c/(ac+c+1)
= a/(ab+a+abc) + b/(bc+b+1) + c/(ac+c+1)
= 1/(b+1+bc) + b/(bc+b+1) + c/(ac+c+1)
= 1/(bc+b+1) + b/(bc+b+1) + c/(ac+c+1)
= (1+b)/(bc+b+1) + c/(ac+c+1)
= (abc+b)/(bc+b+abc) + c/(ac+c+1)
= (ac+1)/(c+1+ac) + c/(ac+c+1)
= (ac+1)/(ac+c+1) + c/(ac+c+1)
= (ac+1+c)/(ac+c+1)
= 1更多10條
已知abc=1，求ab+a+1分之a加上bc+b+1分之b加上ca+c+1分之c的值
14樓：最好的幸福
abc=1
所以b=1/ac
ab=1/c
bc=1/a
所以原式=a/(1/c+a+1)+(1/ac)/(1/a+1/ac+1)+c/(ac+c+1)
第乙個式子上下同乘c
第二個式子上下同乘ac
所以=ac/(ac+c+1)+1/(ac+c+1)+c/(ac+c+1)
=(ac+c+1)/(ac+c+1)=1
已知正數a b c 滿足abc 1 求
a ab a 1 ab abc ab a abc abac abc ab 就是第2個式子乘以a，第3個乘以ab a ab 1 ab a 1 1 a ab a 1 a ab a 1 b bc b 1 ab ab a 1 分子，分母都乘以a，且分母中的abc 1 c ac c 1 1 ab a 1 分子...
已知正整數a，b,c,且a b c 1。求（a
解釋一下樓上a 1 81a 2 9,b 1 81b 2 9,c 1 81c 2 9原因 正數a，b,c,a b c 1，a 1 a 2 b 1 b 2 c 1 c 2可知abc等價，a b c 1 3時取最小值 什麼原理忘了，補奧數時有講 記住，很重要，也可以說是數學感覺 期望，然後向期望努力。是嘗...
已知正數a,b,c，滿足a b c 1，求證（
1 a 1 b 1 c 1 a 1 b 1 c a b c 1 b a c a a b 1 c b a c b c 1 3 b a b a a c c a b c c b 3 2 2 2 9 當a b c時等號成立 1 a 1 1 a a a b c a a b c a b c 2 bc 1 a 1...
已知a,b,c為正實，a b c 1，求證
a b c 2 a 2 b 2 c 2 2ab 2bc 2ac且2ab a 2 b 2 2ac a 2 c 2 2bc b 2 c 2 所以 a 2 b 2 c 2 2ab 2bc 2ac 3 a 2 b 2 c 2 所以 3 a 2 b 2 c 2 1 所以 a 2 b 2 c 2 1 3 p q...
設a,b,c都是正數且a b c 1,求證 1 a
左式 1 a 1 b 1 c a b c a a b c b a b c c a b a c a b b c a c b c 2 a b a c 2 a b b c 2 a c b c 由均值不等式 8 b c a b a c 8 1 a 1 b 1 c 右式 a,b,c都是正數，所以1 a 0，a...