Question

Помогите пожалуйста, даю 80 баллов
С решением

Answer 1

1)

а)

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

б)

$\frac{a {b}^{5} - {b}^{2} }{ {b}^{2} } = \frac{ {b }^{2}(a {b}^{3} - 1) }{ {b}^{2} } = a {b}^{3} - 1$

в)

$\frac{4 {x}^{2} - 4 {y}^{2} }{2x - 2y} = \frac{(2x - 2y)(2x + 2y)}{2x - 2y} = 2x + 2y$

2)

а)

$\frac{3}{a} + \frac{a - 3}{a + 5 {a}^{2} } = \frac{3}{a} + \frac{a - 3}{a(1 + 5a)} = \frac{3(1 + 5a)}{a(1 + 5a)} + \frac{a - 3}{a(1 + 5a)} = \frac{3 + 15a + a - 3}{a(1 + 5a)} = \frac{16a}{a(1 + 5a)} = \frac{16}{1 + 5a}$

б)

$\frac{2 {x}^{2} }{ {x}^{2} -4 } - \frac{2x}{x + 2} = \frac{2 {x}^{2} }{(x - 2)(x + 2)} - \frac{2x}{x + 2} = \frac{2 {x}^{2} }{(x - 2)(x + 2)} - \frac{2x(x - 2)}{(x - 2)(x + 2)} = \frac{2 {x}^{2} - 2 {x}^{2} + 4x}{(x - 2)(x +2 )} = \\ \\ = \frac{4x}{ {x}^{2} - 4}$

в)

$\frac{7a}{a - 2 {b}^{3} } - 7 = \frac{7a}{a - 2 {b}^{3} } - \frac{7(a - 2 {b}^{3}) }{a - 2 {b}^{3} } = \frac{7a - 7a + 14 {b}^{3} }{a - 2 {b}^{3} } = \frac{14 {b}^{3} }{a - 2 {b}^{3} }$

3)

$\frac{9}{ {( a+ 2)}^{2} } - \frac{9}{ {a}^{2} - 4} - \frac{9}{a + 2} = \frac{9}{(a + 2)(a + 2)} - \frac{9}{(a - 2)(a + 2)} - \frac{9}{(a + 2)} = \\ \\ = \frac{9(a - 2)}{(a + 2)(a + 2)(a - 2)} - \frac{9(a + 2)}{(a + 2)(a + 2)(a - 2)} - \frac{9(a + 2)(a - 2)}{(a + 2)(a + 2)(a - 2)} = \\ \\ = \frac{9a - 18}{(a + 2)( {a}^{2} - 4) } - \frac{9a + 18}{(a + 2)( {a}^{2} - 4) } - \frac{9 {a}^{2} - 36}{(a + 2)( {a}^{2} - 4)} = \frac{9a - 18 - 9a - 18 - 9 {a}^{2} + 36 }{(a + 2)( {a}^{2} - 4)} = \\ \\ = \frac{ - 9 {a}^{2} }{(a + 2)( {a}^{2} - 4) } = - \frac{9}{ {a}^{3} + 2 {a }^{2} - 4a - 8 }$

4)

$\frac{2 {a}^{2} - 2 {c}^{2} + {a}^{2}x - {c}^{2} x }{ {x}^{2} - 4} = \frac{2( {a}^{2} - {c}^{2} ) + x( {a}^{2} - {c}^{2} ) }{(x - 2)(x + 2)} = \frac{(x + 2)(a - c)(a + c)}{(x - 2)(x + 2)} = \frac{(a - c)(a + c)}{x - 2} = \\ \\ = \frac{(38.7 - 6.8)(38.7 + 6.8)}{1.9 - 2} = - 14514.5$