Question

помогите это решить очень срочно

Answer 1

$1.a) = {x}^{3}(1 + x) \\ b) = 2( {a}^{2} - 4) = 2(a - 2)(a + 2) \\ c) = x( x + 1) + 2(x + 1) = (x + 2)(x + 1) \\ 2. \: \: \: x(3x - 1) = 0 \\ x = 0 \\ 3x - 1 = 0 \\ x = \frac{1}{3} \\ 5. \: \: \: 2 {x}^{3} + {x}^{2} - 8x - 4 = 0 \\ {x}^{2} (2x + 1) - 4(2x + 1) = 0 \\ ( {x}^{2} - 4)(2x + 1) = 0 \\ {x}^{2} = 4 \\ x = + - 2 \\ 2x + 1 = 0 \\ x = - \frac{1}{2}$
$1.a) = {x}^{4} (x + 1) \\ b) = 2( {a}^{2} - 16) = 2(a + 4)(a - 4) \\ c) = 3x(1 + x) + 8(1 + x) = (3x + 8)(1 + x) \\ 2. \: \: \: x(3x - 4) = 0 \\ x = 0 \\ 3x - 4 = 0 \\ x = - \frac{4}{3} = - 1 \frac{1}{3} \\ 5. \: \: \: {x}^{2} (x - 1) - (x - 1) = 0 \\ ( {x}^{2} - 1)(x - 1) = 0 \\ {x}^{2} = 1 \\ x = - 1 \\ x = 1$