Question

Решите уравнение заранее только на третий на первый и второй есть ответ.​

Answer 1

$\frac{x - 1}{2} + \frac{2x}{3} = \frac{5x}{6} \\ 3(x - 1) + 2 \times 2x = 5x \\ 3x - 3 + 4x = 5x \\ 7x - 5x = 3 \\ 2x = 3 \\ x = \frac{3}{2}$

$\frac{x - 3}{x - 5} + \frac{1}{x} = \frac{x + 5}{x(x - 5)} \\ x \times (x - 3) + x - 5 = x + 5 \\ {x}^{2} - 3x + x - 5 - x - 5 = 0 \\ {x}^{2} - 3x - 10 = 0 \\ D = 9 - 4 \times ( - 10) = 49 \\ \sqrt{D} = 7 \\ x1 = \frac{3 - + 7}{2} = 5 \\ x2 = \frac{3 - 7}{2} = - 2$

$\frac{33 + {x}^{2} }{9 - {x}^{2} } + \frac{7 + x}{x - 3} = - 2 + \frac{4 - x}{x + 3} \\ \frac{33 + {x}^{2} }{(3 - x) (3 + x) } - \frac{7 + x}{3 - x} = - 2 + \frac{4 - x}{x + 3} \\ 33 + {x}^{2} - (7 + x)(3 + x) = - 2(9 - {x}^{2} ) + (4 - x)(3 - x) \\ 33 + {x}^{2} - (21 + 7x + 3x + {x}^{2}) = - 18 + 2 {x}^{2} + 12 - 4x - 3x + {x}^{2} \\ 33 + {x}^{2} - 21 - 10x - {x}^{2} = 6 + 3 {x}^{2} - 7x \\ 12 - 10x - 6 - 3 {x}^{2} + 7x = 0 \\ - 3 {x}^{2} - 3x + 6 = 0( \div (- 3)) \\ {x}^{2} + x - 2 = 0 \\ D = 1 - 4 \times( - 2) = 9 \\ \sqrt{D} = 3 \\ x1 = \frac{ - 1 + 3}{2} = 1 \\ x2 = \frac{ - 1 - 3}{2} = - 2$