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$$\frac{9/x^2}{\large\frac{x^2}{25}+\frac{x^2}{15}}$$

I have attempted to solve this problem a multitude of times and each time I get a different answer. Is it $\frac{360}{x^2}$, $\frac{360}{x^4}$, or as my calculator says $\frac{675}{8x^4}$? Please show the work as well so I understand how it is done. Thanks.

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4 Answers

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Note: answer changed to reflect altered problem statement. $$ \frac{\frac{9}{x^2}}{\frac{x^2}{25}+\frac{x^2}{15}} = \frac{\frac{9}{x^2}}{\frac{(15+25)x^2}{15\cdot 25}} = \frac{9}{x^2\frac{40x^2}{375}} = \frac{375\cdot 9}{40x^4} = \frac{675}{8x^4}. $$

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$$\frac{9/x^2}{\frac{x^2}{25}+\frac{x^2}{15}}=\frac{9/x^2}{\frac{3x^2+5x^2}{75}}=\frac{75*9}{8*x^2*x^2}=\frac{675}{8x^4}$$

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$$\frac{\frac{9}{x^{2}}}{\frac{x^{2}}{25}+\frac{x^{2}}{15}}=\frac{\frac{9}{x^{2}}}{\frac{x^{2}}{25}+\frac{x^{2}}{15}}\times\frac{75x^{2}}{75x^{2}}=\frac{675}{8x^{4}}$$

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First recall that \[ \frac{a}{b} \pm \frac{c}{d} =\frac{ad \pm cb}{bd} \] And \[ \frac{\frac{a}{b}}{\frac{c}{d}} =\frac{ad}{bc} \] Then \[ \frac{\frac{9}{x^{2}}}{\frac{x^{2}}{25}+\frac{x^{2}}{15}}= \frac{\frac{9}{x^{2}}}{\frac{15x^{2}+25x^{2}}{25\cdot 15}}= \frac{\frac{9}{x^{2}}}{\frac{40x^{2}}{375}}=\frac{9\cdot 375}{40x^{4}}= \frac{675}{8x^{4}} \]

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