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Divide the sum of \(\frac{65}{12}\) and \(\frac{8}{3}\) by their differ-renice. |
Answer» According to the question, \((\frac{65}{12}+\frac{8}{3})\div(\frac{65}{12}-\frac{8}{3})\) = \((\frac{65\times1+8\times4}{12})\div(\frac{65\times1-8\times4}{12})\) = \((\frac{65+32}{12})\div(\frac{65-32}{12})\) = \((\frac{97}{12})\div(\frac{33}{12})\) = \(\frac{97}{12}\times\frac{12}{33}\) = \(\frac{97}{33}\) |
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