Question

$\frac{x}{x+1} -\frac{1}{x}=\frac{1}{x^2+x}$
$\frac{x}{x-2}-\frac{1}{x} =\frac{4}{x^2-2x}$
Решите. Дам 35 баллов.

Answer 1

$\frac{x}{x + 1} - \frac{1}{x} = \frac{1}{ {x}^{2} + x} \\ \frac{x}{x + 1} - \frac{1}{x} - \frac{1}{x(x + 1)} = 0 \\ \frac{ {x}^{2} - (x + 1) - 1 }{x(x + 1)} = 0 \\ \frac{ {x}^{2} - x - 1 - 1}{x(x + 1)} = 0 \\ \frac{ {x}^{2} - x - 2}{x(x + 1)} = 0$
ОДЗ: х(х + 1) ≠ 0
х ≠ 0,
х + 1 ≠ 0, х ≠ -1

D = b^2 - 4ac = 1 - 4*(-2) = 9
x1 = (1 + 3)/2 = 2
x2 = (1 - 3)/2 = -1 -- не удовл.

Ответ: 2.

2)
$\frac{x}{x - 2} - \frac{1}{x} = \frac{4}{ {x}^{2} - 2x} \\ \frac{x}{x - 2} - \frac{1}{x} - \frac{4}{x(x - 2)} = 0 \\ \frac{ {x}^{2} - (x - 2) - 4}{x(x - 2)} = 0 \\ \frac{ {x}^{2} - x + 2 - 4 }{x(x - 2)} = 0 \\ \frac{ { x }^{2} - x - 2}{x(x - 2)} = 0$
х(х - 2) ≠ 0
х ≠ 0
х - 2 ≠ 0, х ≠ 2

D = b^2 - 4ac = 1 - 4*(-2) = 9
x1 = (1 + 3)/2 = 2 -- не удовл.
x2 = (1 - 3)/2 = -1

Ответ: -1.