Question

Приведите данные уравнения к квадратным:
1) x(x-3) = 4;
3) 3x(x-5)=x(x+1) - x²;
2) (x-3)(x-1)=12;
4) 7(x² - 1) = 2(x+2)(x-2).​

Answer 1

Объяснение:

$1)x(x - 3) = 4 \\ {x}^{2} - 3x = 4 \\ {x}^{2} - 3x - 4 = 0$

$2)(x - 3)(x - 1) = 12 \\ {x}^{2} - x - 3x + 3 = 12 \\ {x}^{2} - x - 3x + 3 - 12 = 0 \\ {x}^{2} - 4x - 9 = 0$

$3)3x(x - 5) = x(x + 1) - {x}^{2} \\ 3x ^{2} - 15x = {x}^{2} + x - {x}^{2} \\ 3 {x}^{2} - 15x - {x}^{2} + x + {x}^{2} = 0 \\ 3x^{2} - 14x = 0$

$4)7( {x}^{2} - 1) = 2(x + 2)(x - 2) \\ 7x ^{2} - 7 = 2( {x}^{2} - 4) \\ 7x^{2} - 7 = 2 {x}^{2} - 8 \\ 7 {x}^{2} - 7 - 2 {x}^{2} + 8 = 0 \\ 5 {x}^{2} + 1 = 0$