What is (1/a+1/b+1/c)/(1/(ab)+1/(bc)+1/(ac)) ?

The solution to (1/a+1/b+1/c)/(1/(ab)+1/(bc)+1/(ac)) is (bc+ac+ab)/(c+a+b)

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simplify (1/a+1/b+1/c)/(1/(ab)+1/(bc)+1/(ac))

simplify

\frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{\frac{1}{ab} + \frac{1}{b c} + \frac{1}{a c}}

\frac{b c + a c + ab}{c + a + b}

Solution steps

\frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{\frac{1}{ab} + \frac{1}{b c} + \frac{1}{a c}}

Join

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} : \frac{b c + a c + ab}{ab c}

= \frac{\frac{b c + a c + ab}{ab c}}{\frac{1}{ab} + \frac{1}{b c} + \frac{1}{a c}}

Join

\frac{1}{ab} + \frac{1}{b c} + \frac{1}{a c} : \frac{c + a + b}{ab c}

= \frac{\frac{b c + a c + ab}{ab c}}{\frac{c + a + b}{ab c}}

Apply the fraction rule:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{( b c + a c + ab ) ab c}{ab c ( c + a + b )}

Cancel

\frac{( b c + a c + ab ) ab c}{ab c ( c + a + b )} : \frac{b c + a c + ab}{c + a + b}

= \frac{b c + a c + ab}{c + a + b}

4 y^{2} - 5 y - 3 = 0

n^{2} - 6 n - 19 = 0

2 + 7 n = - 4 + 5 n

\frac{1}{3} (9 x + 3) = 3 x + 1

5 (y + \frac{2}{3}) - 3 = 4 (3 y - \frac{1}{2})

What is (1/a+1/b+1/c)/(1/(ab)+1/(bc)+1/(ac)) ?
The solution to (1/a+1/b+1/c)/(1/(ab)+1/(bc)+1/(ac)) is (bc+ac+ab)/(c+a+b)