Math Problem Solver

\[ \frac{1}{72} + \frac{1}{72} + \frac{6}{8} + \frac{1}{9} + \frac{1}{9} = 1 \]

The sum can be computed by rewriting each term over a common denominator, where the least common denominator of \(\frac{1}{72}\), \(\frac{1}{72}\), \(\frac{6}{8}\), \(\frac{1}{9}\), and \(\frac{1}{9}\) is \(72\). \[ \begin{aligned} \frac{1}{72} + \frac{1}{72} + \frac{6}{8} + \frac{1}{9} + \frac{1}{9} &= \frac{1 \cdot 1}{72 \cdot 1} + \frac{1 \cdot 1}{72 \cdot 1} + \frac{6 \cdot 9}{8 \cdot 9} + \frac{1 \cdot 8}{9 \cdot 8} + \frac{1 \cdot 8}{9 \cdot 8} \\ &= \frac{1 + 1 + 54 + 8 + 8}{72} \\ &= \frac{72}{72} \\&= 1 \end{aligned} \]