Решите пж Вариант 1!Нужно хорошее решение!

1.
a)
$\frac{2}{3} + \frac{4}{11} = \frac{22}{33} + \frac{12}{33} = \frac{34}{33} = 1 \frac{1}{33}$
б)
$\frac{3x}{5} - \frac{2y}{7} = \frac{21x}{35} - \frac{10y}{35} = \frac{21x - 10y}{35}$
2.
$\frac{1}{x + 2} + \frac{4}{ {x}^{2} - 4} = \frac{1}{x + 2} + \frac{4}{(x - 2)(x + 2)} = \frac{x - 2}{(x - 2)(x + 2)} + \frac{4}{(x - 2)(x + 2)} = \frac{x - 2 + 4}{(x - 2)(x + 2)} = \frac{x + 2}{(x - 2)(x + 2)} = \frac{1}{x - 2}$

3.
$\frac{a - 3}{ {a}^{2} + 3a + 9} + \frac{9a}{ {a}^{3} - 27 } - \frac{1}{a - 3} = \frac{a - 3}{ {a}^{2} + 3a + 9} + \frac{9a}{(a - 3)( {a}^{2} + 3a + 9)} - \frac{1}{a -3 } = \frac{ {(a - 3)}^{2} }{(a - 3)( {a}^{2} + 3a + 9)} + \frac{9a}{(a - 3)( {a}^{2} + 3a + 9) } - \frac{ {a}^{2} + 3a + 9}{(a - 3)( {a}^{2} + 3a + 9)} = \frac{ {a}^{2} - 6a + 9 + 9a - {a}^{2} - 3a - 9 }{(a - 3)( {a }^{2} + 3a + 9) } = 0$