6. 1 (1)-+ X 1. 4 = 0; x²+3 1 2 (3) x+2+x24x2 2 (5)+ X 3 1 2 x²+x x+1 2 = (7)x=x²-2x

(1) \[\frac{1}{x} + \frac{4}{x^2 + 3} = 0\] \(\frac{1}{x} = -\frac{4}{x^2 + 3}\) \(x^2 + 3 = -4x\) \(x^2 + 4x + 3 = 0\) \((x + 1)(x + 3) = 0\) \(x = -1, -3\)

Проверка:

Для x = -1:\[\frac{1}{-1} + \frac{4}{(-1)^2 + 3} = -1 + \frac{4}{4} = -1 + 1 = 0\] Для x = -3:\[\frac{1}{-3} + \frac{4}{(-3)^2 + 3} = -\frac{1}{3} + \frac{4}{12} = -\frac{1}{3} + \frac{1}{3} = 0\] (3) \(\frac{1}{x + 2} + \frac{1}{x^2 - 4} = \frac{2}{x - 2}\) \(\frac{1}{x + 2} + \frac{1}{(x + 2)(x - 2)} = \frac{2}{x - 2}\) Умножаем обе части на (x + 2)(x - 2):\[(x - 2) + 1 = 2(x + 2)\]\[x - 1 = 2x + 4\]\[x = -5\]

Проверка:

\[\frac{1}{-5 + 2} + \frac{1}{(-5)^2 - 4} = \frac{1}{-3} + \frac{1}{21} = \frac{-7 + 1}{21} = \frac{-6}{21} = -\frac{2}{7}\]\[\frac{2}{-5 - 2} = \frac{2}{-7} = -\frac{2}{7}\] (5) \(\frac{2}{x} + \frac{1}{x^2 + x} = \frac{2}{x + 1}\) \(\frac{2}{x} + \frac{1}{x(x + 1)} = \frac{2}{x + 1}\) Умножаем обе части на x(x + 1):\[2(x + 1) + 1 = 2x\]\[2x + 2 + 1 = 2x\]\[3 = 0\]

Решений нет.

(7) \(\frac{3}{x} = \frac{2}{x^2 - 2x}\) \(\frac{3}{x} = \frac{2}{x(x - 2)}\) Умножаем обе части на x(x - 2):\[3(x - 2) = 2\]\[3x - 6 = 2\]\[3x = 8\]\[x = \frac{8}{3}\]

Проверка:

\[\frac{3}{\frac{8}{3}} = \frac{9}{8}\]\[\frac{2}{(\frac{8}{3})^2 - 2(\frac{8}{3})} = \frac{2}{\frac{64}{9} - \frac{16}{3}} = \frac{2}{\frac{64 - 48}{9}} = \frac{2}{\frac{16}{9}} = \frac{18}{16} = \frac{9}{8}\]