272. Найдите множество решений неравенства: a) 3x² + 40x + 10 < -x² + 11x + 3; б) 9x2 - x + 9> 3x² + 18x – 6; в) 2x² + 8x - 111 < (3x - 5)(2x + 6); г) (5x + 1)(3x - 1) > (4x - 1)(x + 2).

a) 3x² + 40x + 10 < -x² + 11x + 3 $$4x^2 + 29x + 7 < 0$$ $$4x^2 + 29x + 7 = 0$$ $$D = 29^2 - 4*4*7 = 841 - 112 = 729$$ $$x_1 = \frac{-29 + \sqrt{729}}{8} = \frac{-29 + 27}{8} = -\frac{1}{4}$$ $$x_2 = \frac{-29 - \sqrt{729}}{8} = \frac{-29 - 27}{8} = -7$$ So the inequality holds true in the range $$-7 < x < -\frac{1}{4}$$ б) 9x² - x + 9 > 3x² + 18x - 6 $$6x^2 - 19x + 15 > 0$$ $$6x^2 - 19x + 15 = 0$$ $$D = (-19)^2 - 4*6*15 = 361 - 360 = 1$$ $$x_1 = \frac{19+1}{12} = \frac{20}{12} = \frac{5}{3}$$ $$x_2 = \frac{19-1}{12} = \frac{18}{12} = \frac{3}{2}$$ $$x < \frac{3}{2}$$ or $$x > \frac{5}{3}$$ в) 2x² + 8x - 111 < (3x - 5)(2x + 6) $$2x^2 + 8x - 111 < 6x^2 + 18x - 10x - 30$$ $$4x^2 < 8x + 81$$ $$4x^2 + 8x - 81 > 0$$ $$D = 64 + 4*4*81$$ $$D = 64 + 1296$$ $$D = 1360$$ $$D = \sqrt{1360} = 4\sqrt{85}$$ $$x_1 = \frac{-8 + 4\sqrt{85}}{8} = \frac{-2 + \sqrt{85}}{2}$$ $$x_2 = \frac{-8 - 4\sqrt{85}}{8} = \frac{-2 - \sqrt{85}}{2}$$ So the inequality holds true in the ranges $$x<\frac{-2 - \sqrt{85}}{2}$$ or $$x>\frac{-2 + \sqrt{85}}{2}$$ г) (5x + 1)(3x - 1) > (4x - 1)(x + 2) $$15x^2 - 2x - 1 > 4x^2 + 7x - 2$$ $$11x^2 - 9x + 1 > 0$$ $$D = 81 - 4*11*1 = 81 - 44 = 37$$ $$x_1 = \frac{9 + \sqrt{37}}{22}$$ $$x_2 = \frac{9 - \sqrt{37}}{22}$$ So the inequality holds true in the ranges $$x<\frac{9 - \sqrt{37}}{22}$$ or $$x>\frac{9 + \sqrt{37}}{22}$$ Ответ: a) $$-7 < x < -\frac{1}{4}$$, б) $$x < \frac{3}{2}$$ or $$x > \frac{5}{3}$$, в) $$x<\frac{-2 - \sqrt{85}}{2}$$ or $$x>\frac{-2 + \sqrt{85}}{2}$$, г) $$x<\frac{9 - \sqrt{37}}{22}$$ or $$x>\frac{9 + \sqrt{37}}{22}$$