Question-150838

Question Number 150838 by mathdanisur last updated on 15/Aug/21

Answered by amin96 last updated on 15/Aug/21

\frac{(3^{4} + \frac{1}{4}) (5^{4} + \frac{1}{4}) (7^{4} + \frac{1}{4})}{(4^{4} + \frac{1}{4}) (6^{4} + \frac{1}{4}) (8^{4} + \frac{1}{4})} = \frac{({(3^{2} + \frac{1}{2})}^{2} - 9) ({(5^{2} + \frac{1}{2})}^{2} - 25) ({(7^{2} + \frac{1}{2})}^{2} - 49)}{({(4^{2} + \frac{1}{2})}^{2} - 16) ({(6^{2} + \frac{1}{2})}^{2} - 36) ({(8^{2} + \frac{1}{2})}^{2} - 64)} =

= \frac{(9 + \frac{1}{2} - 3) (9 + \frac{1}{2} + 3) (25 + \frac{1}{2} - 5) (25 + \frac{1}{2} + 5) (49 + \frac{1}{2} - 7) (49 + \frac{1}{2} + 7)}{(16 + \frac{1}{2} - 4) (16 + \frac{1}{2} + 4) (36 + \frac{1}{2} + 6) (36 + \frac{1}{2} - 6) (64 + \frac{1}{2} - 8) (64 + \frac{1}{2} + 8)} =

= \frac{\frac{13}{2} \cdot \frac{25}{2} \cdot \frac{41}{2} \cdot \frac{61}{2} \cdot \frac{85}{2} \cdot \frac{113}{2}}{\frac{25}{2} \cdot \frac{41}{2} \cdot \frac{85}{2} \cdot \frac{61}{2} \cdot \frac{113}{2} \cdot \frac{145}{2}} = \frac{13}{145}

Commented by mathdanisur last updated on 15/Aug/21

Ser thank you