Question

АЛГЕБРА 7 КЛАСС! ДАЮ 35 БАЛОВ! СРОЧНО!!! РЕШИТЬ 2 ПРИМЕРА!

Answer 1

1.

$\frac{x + 3}{x - 3} - \frac{x - 3}{x + 3} = \frac{12x}{ {x}^{2} - 9 } \\ \frac{(x + 3)(x + 3)}{(x - 3)(x + 3)} - \frac{(x - 3)(x - 3)}{(x - 3)(x + 3)} = \frac{12x}{ {x}^{2} - 9} \\ \frac{ {(x + 3)}^{2} - {(x - 3)}^{2} }{ {x}^{2} - 9} = \frac{12x}{ {x}^{2} - 9} \\ \frac{(x + 3 - x + 3)(x + 3 + x - 3)}{ {x}^{2} - 9} = \frac{12x}{ {x}^{2} - 9} \\ \frac{6 \times 2x}{ {x}^{2} - 9 } = \frac{12x}{ {x}^{2} - 9 } \\ \frac{12x}{ {x}^{2} - 9} = \frac{12x}{ {x}^{2} - 9}$

2.

$\frac{6}{ {x}^{2} - 1 } + \frac{3}{x - 1} = \frac{6}{ {x}^{2} - 1 } + \frac{3(x + 1)}{ {x}^{2} - 1 } = \frac{6 + 3x + 3}{ {x}^{2} - 1} = \frac{9 + 3x}{ {x}^{2} - 1}$

при х = 20:

$\frac{9 + 60}{400 - 1} = \frac{69}{399}$