Question

решите пж 1 и 2 вариант даю 100 балов​

Answer 1

1 вариант

1)

а)

$\frac{2}{3} + \frac{4}{11} = \frac{22}{33} + \frac{12}{33} = \frac{34}{33} = 1 \frac{1}{33}$

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.

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

2 вариант

1.

а)

$\frac{1}{7} + \frac{3}{14} = \frac{2}{14} + \frac{3}{14} = \frac{5}{14}$

б)

$\frac{2x}{3} - \frac{3y}{5} = \frac{10x}{15} - \frac{9y}{15} = \frac{10x - 9y}{15}$

2.

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

3.

$\frac{a - 4}{ {a}^{2} + 4a + 16} + \frac{12a}{ {a}^{3} - 64 } - \frac{1}{a - 4} = \frac{a - 4}{ {a}^{2} + 4a + 16 } + \frac{12a}{(a - 4)( {a}^{2} + 4a + 16) } - \frac{1}{a - 4} = \\ \\ = \frac{ {(a - 4)}^{2} }{(a - 4)( {a}^{2} + 4a + 16) } + \frac{12a}{(a - 4)( {a }^{2} + 4a + 16)} - \frac{ {a}^{2} + 4a + 16 }{(a - 4)( {a}^{2} + 4a + 16) } = \frac{ {a}^{2} - 8a + 16 + 12a - {a}^{2} - 4a - 16 }{(a - 4)( {a}^{2} + 4a + 16} = \\ \\ = \frac{0}{ {a}^{3} - 64 } = 0$