Question

ПОЖАЛУЙСТА НАПИШИТЕ НА ЛИСТОЧКЕ!!!

Answer 1

$a) \frac{3a}{a + 1} + \frac{3}{a + 1} = \frac{3a + 3}{a + 1} = \frac{3(a + 1)}{a + 1} = 3$

$б) \frac{2c - 1}{4c + 2} + \frac{2c + 1}{6c - 3} + \frac{4c}{4c {}^{2} - 1} = \frac{2c - 1}{2(2c + 1)} + \frac{2c + 1}{3(2c - 1)} + \frac{4c}{(2c - 1) \times (2c + 1)} = \frac{3(2c - 1) {}^{2} + 2(2c + 1) {}^{2} + 24c }{6(2c - 1) \times (2c + 1)} = \frac{3(4c {}^{2} - 4c + 1) + 2(4c {}^{2} + 4c + 1) + 24c}{6(4c {}^{2} - 1) } = \frac{12c {}^{2} - 12c + 3 + 8c {}^{2} + 8c + 2 + 24c }{6(2c - 1) \times (2c + 1)} = \frac{20c {}^{2} + 20c + 5 }{6(2c - 1) \times (2c + 1)} = \frac{5(4c {}^{2} + 4c + 1) }{6(2c - 1) \times (2c + 1)} = \frac{5(2c + 1) {}^{2} }{6(2c - 1) \times (2c + 1)} = \frac{5(2c + 1)}{6(2c - 1)} = \frac{10c + 5}{12c - 6}$