Question

СРОЧНО ПОЖАЛУЙСТА ПОМОГИТЕ РЕШИТЬ!!!​

Answer 1

1)

$\frac{ - 1 - 5}{ {( - 1)}^{2} + 5 } = \frac{ - 6}{6} = - 1$

Б) -1

2)

$\frac{ {x}^{12} }{ {x}^{6} } = {x}^{6}$

Б)

3)

$\frac{2}{x - 1} + \frac{x + 7}{x - 1} = \frac{2 + x - 7}{x - 1} = \frac{x - 5}{x - 1}$

А)

4)

${a}^{2} - 36 = 0 \\ (a - 6)(a + 6) = 0 \\ a = 6 \\ a = - 6$

В)

5)

$\frac{2m}{m - n} + \frac{2n}{n - m} = \frac{2m}{m - n} - \frac{2n}{m - n} = \frac{2m - 2n}{m - n} = \frac{2(m - n)}{m - n} = 2$

Б)

6)

$\frac{6 {x}^{2} + y }{3x} - 2x = \frac{6 {x}^{2} + y - 6 {x}^{2} }{3x} = \frac{y}{3x}$

Б

7)

$\frac{ {a}^{3} - {a}^{5} }{ {a}^{3} - a } = \frac{ - {a}^{2}( {a}^{3} - a)}{ {a}^{3} - a } = - {a}^{2}$

$\frac{xy - x {y}^{2} }{ {y}^{2} - 1 } = \frac{ - xy( - 1 + y)}{(y - 1)( y + 1)} = - \frac{xy}{y + 1}$

8)

$\frac{3}{a - 2} - \frac{a + 3}{4 - {a}^{2} } + \frac{3a + 1}{ {a}^{2} - 4a + 4} = \frac{3}{a - 2} + \frac{a + 3}{(a - 2)(a + 2)} + \frac{3a + 1}{ {(a - 2)}^{2} } = \frac{3( {a}^{2} - 4) + (a + 3)(a - 2) + (3a + 1)(a + 2) }{ {(a - 2)}^{2}(a + 2) } = \frac{3 {a}^{2} - 12 + {a}^{2} - 2a + 3a - 6 + 3 {a}^{2} + 6a + a + 2 }{ {(a - 2)}^{2}(a + 2) } = \frac{7 {a}^{2} + 8a - 16 }{ {(a - 2)}^{2} (a + 2)}$