Question

Решите а, б. даю 20 баллов номер 793

Answer 1

Задание 793

а)

$\frac{2x-3}{x+5} + \frac{3x+1}{x - 5} = \frac{(2x-3)(3x+1)}{(x+5)(x-5)}\\\\ODZ:\\\\x\neq5;\\\\ x\neq-5\\\\(2x - 3)(x-5) + (3x+1)(x+5) = (2x-3)(3x+1)\\\\2x^2-10x-3x+15 +3x^2+15x+x+5 = 6x^2+2x -9x -3\\\\2x^2-13x+15 +3x^2+16x+5 - 6x^2+7x + 3=0\\\\2x^2 + 3x^2 - 6x^2 - 13x + 16x +7x + 15 + 5 + 3 = 0\\\\-x^2 + 10x + 23 = 0 |: (-1)\\\\x^2 - 10x - 23 = 0\\\\D = 100 + 92 = 192\\\\\sqrt{D} = \sqrt{192} = 8\sqrt{3} \\\\ x_1 = \frac{10-8\sqrt{3}}{2} = 5 - 4\sqrt{3}\\\\x_2 = \frac{10+8\sqrt{3}}{2} = 5 + 4\sqrt{3}$

б)

$\frac{4x+1}{x-2} - \frac{2x-5}{x+3} = \frac{4x-47}{x^2+x -6}\\\\\frac{4x+1}{x-2} - \frac{2x-5}{x+3} = \frac{4x-47}{(x-2)(x+3)}\\\\ODZ:\\\\x\neq2\\\\x\neq-3\\\\(4x - 1)(x + 3) - (2x-5)(x-2) = 4x - 47\\\\4x^2 + 12x - x- 3 -2x^2 + 4x + 5x - 10 = 4x - 47\\\\4x^2 - 2x^2 + 11x + 9x - 4x - 3 - 10 + 47 = 0\\\\2x^2 + 16x + 34 = 0|:2\\\\x^2 + 8x + 17 = 0\\\\D = 64 - 78 = -4$

У этого уравнения нет решений на множестве действительных чисел $R$.