Question

2.136. Найдите корни уравнения (1,2)​

Answer 1

1)

$\dfrac{x-4}{x-5} + \dfrac{x-6}{x-5} = 2\\\\x - 4 + x - 6 = 2(x-5), \; x\neq 5\\\\2x - 10 = 2x - 10\\\\x \in (-\infty; 5) \cup (5; +\infty)$

2)

$\dfrac{1}{2-x}-1=\dfrac{1}{x-2}-\dfrac{6-x}{3x^2-12}\\\\\dfrac{6-x}{3x^2-12}-\dfrac{1}{x-2}+\dfrac{1}{2-x}-1=0\\\\\dfrac{6-x}{3(x-2)(x+2)}-\dfrac{3(x+2)}{3(x-2)(x+2)} - \dfrac{3(x+2)}{3(x-2)(x+2)} - \dfrac{3(x-2)(x+2)}{3(x-2)(x+2)}=0\\\\\dfrac{6-x-3x-6-3x-6-3x^2+12}{3(x-2)(x+2)}=0\\\\\dfrac{-3x^2-7x+6}{3(x-2)(x+2)}=0\\\\-3x^2-7x+6=0, \; x \neq \pm2\\\\D = 49 + 4\cdot3\cdot6 = 121\\\\x_1 = \dfrac{7-\sqrt{121}}{-2\cdot3}=\dfrac{2}{3}\\\\x_2 = \dfrac{7+\sqrt{121}}{-2\cdot3}=-3$