22 dez 2012, 19:08
\(Se\ x\ =\log_{c}(ab)\ ,\ y=\log_{b}(ac)\ e\ z\ = \log_{a}(bc) \ prove\ que:\\ \\ 1/x+1 \ +\ 1/y+1\ + 1/z+1\ =\ 1.\)
23 dez 2012, 16:37
É isto
\(1/x+1 \ +\ 1/y+1\ + 1/z+1\ =\ 1\)
OU
isto
\(1/(x+1) \ +\ 1/(y+1)\ + 1/(z+1)\ =\ 1\)
???
23 dez 2012, 20:30
...
Editado pela última vez por
EAFO em 23 dez 2012, 20:35, num total de 1 vez.
23 dez 2012, 20:34
É o segundo caso:
1/(x+1) + 1/(y+1) + 1/(z+1) = 1
25 dez 2012, 21:51
\(\frac{1}{1+\log_{c}(ab)}+\frac{1}{1+\log_{b}(ac)}+\frac{1}{1+\log_{a}(bc)}=1\)
lembre-se que
\(log_{y}(x)=\frac{ln(x)}{ln(y)}\)
\(log(a.b)=log(a)+log(b)\)
então
\(\frac{1}{1+\log_{c}(ab)}+\frac{1}{1+\log_{b}(ac)}+\frac{1}{1+\log_{a}(bc)}=1\\ \frac{1}{1+\frac{\ln(ab)}{\ln(c)}}+\frac{1}{1+\frac{\ln(ac)}{\ln(b)}}+\frac{1}{1+\frac{\ln(bc)}{\ln(a)}}=1\\ \frac{1}{1+\frac{\ln(a)+\ln(b)}{\ln(c)}}+\frac{1}{1+\frac{\ln(a)+\ln(c)}{\ln(b)}}+\frac{1}{1+\frac{\ln(b)+\ln(c)}{\ln(a)}}=1\\ \frac{\ln(c)}{\ln(c)+\ln(a)+\ln(b)}+\frac{\ln(b)}{\ln(b)+\ln(a)+\ln(c)}+\frac{\ln(a)}{\ln(b)+\ln(b)+\ln(c)}=1\)
logo
\(\frac{\ln(c)+\ln(b)+\ln(a)}{\ln(a)+\ln(b)+\ln(c)}=1\)
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