Question

Решите пж Вариант 1!
Нужно хорошее решение!

Answer 1

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

Answer 2

Спасибо