Question

Решите пжжж 4 и 5 номер пж! НАДО ОЧЕНЬ СРОЧНО!

Answer 1

№4
$a) \: \frac{n + 3}{2n + 2} - \frac{n + 1}{2n - 2} + \frac{3}{ {n}^{2} - 1 } = \frac{n + 3}{2(n + 1)} - \frac{n + 1}{2(n - 1)} + \frac{3}{(n - 1)(n + 1) } = \frac{(n + 3)(n - 1) - (n + 1)(n + 1) + 3 \times 2}{2( {n}^{2} - 1) } = \frac{ {n}^{2} - 2n - 3 - {n}^{2} - 2n - 1 + 6}{2( {n}^{2} - 1) } = \frac{ - 4n + 2}{2( {n}^{2} - 1)} = \frac{ - 2(2n - 1)}{ 2( {n}^{2} - 1) } = \frac{1 - 2n}{{n}^{2} - 1}$
$b) \frac{2 {a}^{2} + 7a + 9 }{ {a}^{3} - 1} + \frac{4a + 3}{ {a}^{2} + a + 1} = \frac{2 {a}^{2} + 7a + 9 }{ (a - 1)({a}^{2} + a + 1)} + \frac{4a + 3}{ {a}^{2} + a + 1} = \frac{2 {a}^{2} + 7a + 9 + (4a + 3)(a - 1) }{(a - 1)({a}^{2} + a + 1)} = \frac{2 {a}^{2} + 7a + 9 +4 {a}^{2} - a - 3 }{(a - 1)({a}^{2} + a + 1)} = \frac{6 {a}^{2} + 6a + 6}{(a - 1)({a}^{2} + a + 1)} = \frac{6({a}^{2} + a + 1)}{(a - 1)({a}^{2} + a + 1)} = \frac{6}{a - 1}$
№5
$\frac{a - 5b}{b} = 9 \\ \frac{a}{b} - 5 = 9 \\ a) \frac{a}{b} = 14 \\ b) \frac{3a + b}{a} = 3 + \frac{b}{a} = 3 + \frac{b}{14b} = 3 + \frac{1}{14} = 3 \frac{1}{14}$