Question

помогите это решить очень срочно

Answer 1

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