Question

Помогите пожалуйста с 1-4

Answer 1

1)
$\frac{ {a}^{2} + 1 }{ {a}^{2} - 2a + 1} + \frac{a + 1}{a - 1} = \frac{ {a}^{2} + 1 }{ {(a - 1)}^{2} } + \frac{a + 1}{a - 1} = \frac{ {a}^{2} + 1}{ {(a - 1)}^{2} } + \frac{(a + 1)(a - 1)}{ {(a - 1)}^{2} } = \frac{ {a}^{2} + 1 + {a}^{2} - 1}{ {(a - 1)}^{2} } = \frac{2 {a}^{2} }{ {(a - 1)}^{2} }$
2)
$\frac{ {a}^{2} + {b}^{2} }{ {a}^{2} - {b}^{2} } - \frac{a - b}{a + b} = \frac{ {a}^{2} + {b}^{2} }{ {a}^{2} - {b}^{2} } - \frac{(a - b)^{2} }{(a + b)(a - b)} = \frac{ {a}^{2} + {b}^{2} }{ {a}^{2} - {b}^{2} } - \frac{ {(a - b)}^{2} }{ {a}^{2} - {b}^{2} } = \frac{ {a}^{2} + {b}^{2} - {a}^{2} + 2ab - {b}^{2} }{ {a}^{2} - {b}^{2} } = \frac{2ab}{ {a}^{2} - {b}^{2} }$
3)
$\frac{c + 7}{c - 7} + \frac{28c}{49 - {c}^{2} } = \frac{c + 7}{c - 7} - \frac{28c}{ {c}^{2} - 49 } = \frac{ {(c + 7)}^{2} }{(c - 7)(c + 7)} - \frac{28c}{ (c - 7)(c + 7)} = \frac{ {c}^{2} + 14c + 49 - 28c}{(c - 7)(c + 7)} = \frac{ {c}^{2} - 14c + 49}{(c - 7)(c + 7)} = \frac{ {(c - 7)}^{2} }{(c - 7)(c + 7)} = \frac{c - 7}{c + 7}$
4)
$\frac{5a + 3}{2 {a}^{2} + 6a } + \frac{6 - 3a}{ {a}^{2} - 9 } = \frac{5a + 3}{2a(a + 3)} + \frac{6 - 3a}{(a - 3)(a + 3)} = \frac{(5a + 3)(a - 3) + (6 - 3a) \times 2a}{2a(a + 3)(a - 3)} = \frac{5 {a}^{2} + 3a - 15a - 9 + 12a - 6 {a}^{2} }{2a(a + 3)(a - 3)} = \frac{ - {a}^{2} - 9}{2a(a + 3)( a - 3) }$