1. 5,(41) = 5 + 41/99 = 536/99
2. подставим 1 в многочлен: 5-4+3+9=21 ≠0. Ответ: нет
3. ![\frac{x^{2/10}*x^{10/4}*x^8}{x^{99/3}*x^(2/66)} = \frac{x^{0.2+2.5+8}}{x^{33+1/33}} = x^{10.7-33-1/33} = x^{-22.3-1/33} = x ^ {-(7359+10)/330} = 1/\sqrt[330]{7369}\\ \frac{x^{2/10}*x^{10/4}*x^8}{x^{99/3}*x^(2/66)} = \frac{x^{0.2+2.5+8}}{x^{33+1/33}} = x^{10.7-33-1/33} = x^{-22.3-1/33} = x ^ {-(7359+10)/330} = 1/\sqrt[330]{7369}\\](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%2F10%7D%2Ax%5E%7B10%2F4%7D%2Ax%5E8%7D%7Bx%5E%7B99%2F3%7D%2Ax%5E%282%2F66%29%7D%20%3D%20%20%5Cfrac%7Bx%5E%7B0.2%2B2.5%2B8%7D%7D%7Bx%5E%7B33%2B1%2F33%7D%7D%20%3D%20x%5E%7B10.7-33-1%2F33%7D%20%3D%20x%5E%7B-22.3-1%2F33%7D%20%3D%20x%20%5E%20%7B-%287359%2B10%29%2F330%7D%20%3D%201%2F%5Csqrt%5B330%5D%7B7369%7D%5C%5C)
4А. 
4Б.
(1/3)^{-2}\\ 3x-9>9\\ x>6." alt="3x-9>(1/3)^{-2}\\ 3x-9>9\\ x>6." align="absmiddle" class="latex-formula">
5А. Неопределенность ∞/∞. По правилу Лопиталя 
5Б. Неопределенность ∞/∞. По правилу Лопиталя 
5В. Неопределенность ∞/∞. По правилу Лопиталя 