Question

пж поможіть дам 35 балів​

Answer 1

$67.$

$a) \: \: \: \: \: \frac{ {x}^{3} + 8 }{x + 2} = \frac{(x + 2)( {x}^{2} - 2x +4) }{x + 2} = {x}^{2} - 2x + 4$

$b) \: \: \: \: \: \frac{ {z}^{2} + 3z + 9 }{27 - {z}^{3} } = \frac{ {z}^{2} + 3z + 9 }{(3 - z)(9 + 3z + {z}^{2} )} = \frac{1}{3 - z}$

$c) \: \: \: \: \: \frac{ {y}^{6} - 1}{1 - {y}^{2} } = \frac{( {y}^{2} - 1)( {y}^{4} + {y}^{2} + 1) }{ - ( {y}^{2} - 1)} = - ({y}^{4} + {y}^{2} + 1) = - {y}^{4} - {y}^{2} - 1$

$68.$

$a) \: \: \: \: \: \frac{ax + cx - ay - cy}{cx - cy} = \frac{x(a + c) - y(a + c)}{c(x - y)} = \frac{(a + c)(x - y)}{c(x - y)} = \frac{a + c}{c}$

$b) \: \: \: \: \: \frac{ {b}^{2} + 2ab + {a}^{2} }{ {a}^{2} + ab - ax - bx} = \frac{ {(a + b)}^{2} }{a(a + b) - x(a + b)} = \frac{ {(a + b)}^{2} }{(a + b)(a - x)} = \frac{a + b}{a - x}$

$c) \: \: \: \: \: \frac{8a + 4b}{2ab + {b}^{2} - 2ad - bd} = \frac{4(2a + b)}{b(2a + b) - d(2a + b)} = \frac{4(2a + b)}{(2a + b)(b - d)} = \frac{4}{b - d}$