(algebra)Produtos notáveis

por **Man Utd** » Seg Abr 15, 2013 20:42

e $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=\frac{7}{10}$ , o valor de $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}$ é igual a:
gabarito :39/10

por **e8group** » Seg Abr 15, 2013 21:28

Por

Temos :

Daí ,

$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{b+a} = \frac{7 - (b+c)}{b+c} + \frac{7 - (a+c)}{a+c} + \frac{7 - (a+b)}{b+a}$

$\frac{7}{b+c} - \frac{b+c}{b+c} + \frac{7}{a+c} - \frac{a+c}{a+c} + \frac{7}{a+c} - \frac{a+b}{a+b}=$

$\frac{7}{b+c} + \frac{7}{a+c} + \frac{7}{a+b} - 1 - 1 - 1$ .

Tente concluir .

por **Man Utd** » Ter Abr 16, 2013 09:15

$\\\\ 7.(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})-3 \\\\ 7.(\frac{7}{10})-3 \\\\ \frac{49}{10}-3 \\\\ \frac{19}{10}$

aonde errei? :oops:

por **DanielFerreira** » Ter Abr 16, 2013 11:53

Man Utd,
Não errou! O gabarito está errado.
Fiz de outra forma e conclui o mesmo, veja:

$\\ (a + b + c)\left ( \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{a + c} \right ) = 7 \cdot \frac{7}{10} \\\\\\ \frac{a}{a + b} + \frac{a}{b + c} + \frac{a}{a + c} + \frac{b}{a + b} + \frac{b}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} + \frac{c}{b + c} + \frac{c}{a + c} = \frac{49}{10} \\\\\\ \left (\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right ) + \frac{a}{a + b} + \frac{a}{a + c} + \frac{b}{a + b} + \frac{b}{b + c} + \frac{c}{b + c} + \frac{c}{a + c} = \frac{49}{10} \\\\\\ \left (\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right ) + \frac{a}{a + b} + \frac{b}{a + b} + \frac{a}{a + c} + \frac{c}{a + c} + \frac{b}{b + c} + \frac{c}{b + c} = \frac{49}{10} \\\\\\ \left (\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right ) + \frac{1}{a + b}(a + b) + \frac{1}{a + c}(a + c) + \frac{1}{b + c}(a + b) = \frac{49}{10}$

$\\ \left (\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right ) + 1 + 1 + 1 = \frac{49}{10} \\\\\\ \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} = \frac{49}{10} - 3 \\\\\\ \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} = \frac{49}{10} - \frac{30}{10} \\\\\\ \boxed{\boxed{\boxed{\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} = \frac{19}{10}}}}$