Question

Решите пж последние два задание.
Буду очень благодарна. Если можно побыстрей)))

Answer 1

$( \frac{6}{ {y}^{2} - 9 } + \frac{1}{3 - y} ) \times \frac{ {y}^{2} + 6y + 9}{5} = ( \frac{6}{(y - 3)(y + 3)} - \frac{1}{y - 3} ) \times \frac{ {(y + 3)}^{2} }{5} = \frac{6 - (y + 3)}{(y - 3)(y + 3)} \times \frac{ {(y + 3)}^{2} }{5} = - \frac{(y - 3) {(y + 3)}^{2} }{5(y - 3)(y + 3)} = \frac{y + 3}{5}$
$\frac{3x}{x + 3} - \frac{42}{ {x}^{2} - 9} = 1 + \frac{7}{3 - x} \\ \frac{3x}{x + 3} - \frac{42}{(x - 3)(x + 3)} = 1 - \frac{7}{x - 3} \\ \frac{3x(x - 3) - 42}{(x - 3)(x + 3)} = \frac{x - 3 - 7}{x - 3} \\ \frac{3 {x}^{2} - 9x - 42 }{(x - 3)(x + 3)} - \frac{x - 10}{x - 3} = 0 \\ \frac{3 {x}^{2} - 9x - 42 - (x - 10)(x + 3) }{(x - 3)(x + 3)} = 0 \\ \frac{3 {x}^{2} - 9x - 42 - {x}^{2} - 3x + 10x + 30 }{(x - 3)(x + 3)} = 0 \\ \frac{2 {x}^{2} - 2x - 12 }{(x - 3)(x + 3)} = 0$