Question

Ученикам 8 класса ВОПРОС 8

Answer 1

=
$\frac{81x {}^{2} }{9(9x {}^{2} - 1) } + \frac{54x}{3(3x - 1) \times 3(3x + 1)} + \frac{9}{9(9x {}^{2} - 1)} = \frac{9x {}^{2} }{9x {}^{2} - 1) } + \frac{6x}{(3x - 1)(3x + 1)} + \frac{1}{9x {}^{2} - 1 } = \frac{9x {}^{2} + 6x + 1}{9x - 1} = \frac{(3x +1) {}^{2} }{(3x - 1)(3x + 1)} = \frac{3x + 1}{3x - 1}$вместо t там x