Найдите корни уравнения (2x+1) ^2=2x+1 (y-2) ^2-4=0 (3x-1) ^2=2 (3x-1) 9 - (2x-3) ^2=0

1) (2 * x + 1) ^2 = 2 * x + 1;

4 * х^2 + 2 * х + 2 * х + 1 = 2 * x + 1;

4 * х^2 + 2 * х = 0;

х * (4 * х + 2) = 0;

х1 = 0; х2 = - 2/4 = - 1/2;

2) (y - 2) ^2 - 4 = 0;

у^2 - 4 * у + 4 - 4 = 0;

у * (у - 4) = 0;

у1 = 0; у2 = 4;

3) (3 * x - 1) ^2 = 2 * (3 * x - 1);

9 * х^2 - 3 * х - 3 * х + 1 = 6 * х - 2;

3 * х^2 - 4 * х + 1 = 0;

D = 16 - 12 = 4;

Х1 = (4 + 2) / 6 = 1;

Х2 = (4 - 2) / 6 = 1/3;

4) 9 - (2 * x - 3) ^2 = 0;

9 - 4 * х^2 + 12 * х - 9 = 0;

-х^2 + 3 * х = 0;

х * (-х + 3) = 0;

х1 = 0; х2 = 3.