Выполни действия: 1) -54a⁶b⁹/c¹² * (-c²⁰/12a⁴b¹⁵) 2) 98m⁸/p¹⁷ : (49m⁵p²) 3) x²-49/3x-24 * 5x+35/x-8 4) 5y+20/y²-16y+64 * 6y-48/y+4

1) $$- \frac{54a^6b^9}{c^{12}} \cdot (-\frac{c^{20}}{12a^4b^{15}}) = \frac{54a^6b^9c^{20}}{12a^4b^{15}c^{12}} = \frac{9a^2c^8}{2b^6}$$ 2) $$\frac{98m^8}{p^{17}}:(49m^5p^2) = \frac{98m^8}{p^{17}} \cdot \frac{1}{49m^5p^2} = \frac{2m^3}{p^{19}}$$ 3) $$\frac{x^2-49}{3x-24} \cdot \frac{5x+35}{x-8} = \frac{(x-7)(x+7)}{3(x-8)} \cdot \frac{5(x+7)}{x-8} = \frac{5(x-7)(x+7)^2}{3(x-8)^2}$$ 4) $$\frac{5y+20}{y^2-16y+64} \cdot \frac{6y-48}{y+4} = \frac{5(y+4)}{(y-8)^2} \cdot \frac{6(y-8)}{y+4} = \frac{30}{y-8}$$ 2. 1) $$(\frac{4x}{7y})^3 = \frac{64x^3}{343y^3}$$ 2) $$(\frac{2a^5}{3b^2})^5 = \frac{32a^{25}}{243b^{10}}$$ 3. $$(\frac{-3x^4}{y^7})^5 : (\frac{9x^6}{y^8})^3 = \frac{-243x^{20}}{y^{35}} : \frac{729x^{18}}{y^{24}} = \frac{-243x^{20}}{y^{35}} \cdot \frac{y^{24}}{729x^{18}} = \frac{-x^2}{3y^{11}}$$ 4. 1) $$\frac{x^3+125}{x^2-12x+36} \cdot \frac{x^2-36}{x^2-5x+25} \cdot \frac{11x+66}{x-6} = \frac{(x+5)(x^2-5x+25)}{(x-6)^2} \cdot \frac{(x-6)(x+6)}{x^2-5x+25} \cdot \frac{11(x+6)}{x-6} = 11(x+5)(x+6)^2/(x-6)$$ 2) $$(\frac{a+4}{a-4} - \frac{a-4}{a+4}) : \frac{48a}{16-a^2} = (\frac{(a+4)^2 - (a-4)^2}{(a-4)(a+4)}) : \frac{48a}{16-a^2} = \frac{a^2+8a+16 - (a^2-8a+16)}{(a-4)(a+4)} : \frac{48a}{16-a^2} = \frac{16a}{(a-4)(a+4)} \cdot \frac{-(a-4)(a+4)}{48a} = -\frac{1}{3}$$ 5. $$\frac{\frac{a^2}{a+5} - \frac{a^3}{a^2+10a+25}}{\frac{a}{a+5} - \frac{a^2}{a^2-25}} = \frac{\frac{a^2}{a+5} - \frac{a^3}{(a+5)^2}}{\frac{a}{a+5} - \frac{a^2}{(a-5)(a+5)}} = \frac{\frac{a^2(a+5) - a^3}{(a+5)^2}}{\frac{a(a-5)-a^2}{(a+5)(a-5)}} = \frac{\frac{a^3+5a^2 - a^3}{(a+5)^2}}{\frac{a^2-5a-a^2}{(a+5)(a-5)}} = \frac{\frac{5a^2}{(a+5)^2}}{\frac{-5a}{(a+5)(a-5)}} = \frac{5a^2}{(a+5)^2} \cdot \frac{(a+5)(a-5)}{-5a} = \frac{a(a-5)}{-(a+5)} = -\frac{a^2-5a}{a+5} = \frac{5a-a^2}{a+5}$$ 6. Дано: $$16x^2 + \frac{9}{x^2} = 145$$. Найти: $$4x + \frac{3}{x}$$. Пусть $$y = 4x + \frac{3}{x}$$. Тогда $$y^2 = (4x + \frac{3}{x})^2 = 16x^2 + 2 \cdot 4x \cdot \frac{3}{x} + \frac{9}{x^2} = 16x^2 + 24 + \frac{9}{x^2}$$. Выразим $$16x^2 + \frac{9}{x^2}$$ через $$y^2$$: $$16x^2 + \frac{9}{x^2} = y^2 - 24$$. Подставим это в исходное уравнение: $$y^2 - 24 = 145$$. Решим уравнение относительно $$y$$: $$y^2 = 145 + 24 = 169$$. $$y = \pm \sqrt{169} = \pm 13$$. **Ответ: $$4x + \frac{3}{x} = \pm 13$$**