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Why does `evalb(((3*x + 1)^5)(3*x - 1)/x^6 = ((3 + 1/x)^5)(3 - 1/x)) assuming (0 < x)` return `fa
| Author | Votes | Δ |
|---|---|---|
| acer | 2 | 0.00 |
Last visited: Sep 5, 2020, 12:36:14 PM