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Suppose $a+b+c = ab$, and $a,b,c$ are nonzero. Then

$$\frac{ab + ac + bc + c^2}{abc}$$ is equal to what? I've found an answer: $\frac ba$.

MPW
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Physicer
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  • Well, even assuming that for some specific set of numerical values of $a,b,c$ and $d$, that that's true, what's the question? – Sharat V Chandrasekhar Jun 01 '17 at 21:37
  • Your answer would be incorrect, generally. If you simplify termwise you will get $$\frac1c + \frac1b + \frac 1a + \frac{c}{ab}$$ if that is useful. It certainly depends on the value of $c$. – MPW Jun 01 '17 at 21:40
  • Oh, now I see that you have put some extra conditions in the title of the question -- probably not a good idea. I'm going to add them into the body of the question. – MPW Jun 01 '17 at 21:47

5 Answers5

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$$\frac{ab + (ac + bc + c^2)}{abc}$$ $$=\frac{ab}{abc}+\frac{ c(a + b + c)}{abc}$$ $$=\frac{1}{c} + 1$$

Jay Zha
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For a start: the expression is $$\frac{ab+c(a+b+c)}{abc}=\frac{ab+abc}{abc}=\frac{1}{c}+1 $$

Michael Hoppe
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$$\frac { ab+ac+bc+c^{ 2 } }{ abc } =\frac { a\left( b+c \right) +c\left( b+c \right) }{ abc } =\frac { \left( a+c \right) \left( b+c \right) }{ abc } =\\ =\frac { \left( a+c \right) \left( ab-a \right) }{ abc } =\frac { \left( a+c \right) \left( b-1 \right) }{ bc } =\frac { \left( ab-b \right) \left( b-1 \right) }{ bc } =\\ =\frac { b\left( a-1 \right) \left( b-1 \right) }{ bc } =\frac { ab-a-b+1 }{ c } =\frac { a+b+c-a-b+1 }{ c } =1+\frac { 1 }{ c } $$

haqnatural
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$$a+b+c=ab,\quad a,b,c \in\mathbb{R},\quad a,b,c\neq 0$$

We must simplify the expression $$f(a,b,c)= \frac{ab+ac+bc+c^2}{abc}$$

$$f=\frac{ab+ac+bc+c^2}{(a+b+c)c} =\frac{ab+(a+b+c)c}{(a+b+c)c} =\frac{(a+b+c)+(a+b+c)c}{(a+b+c)c}$$

$$f = \frac{1+c}{c}$$

mrnovice
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$ac+bc+c^2=(a+b+c)c=abc $

So $\frac {ab+ac+bc+c^2}{abc}=$

$\frac {ab+abc}{abc}=$

$\frac 1c +1$.

fleablood
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