Решите уравнение :а) (x-1)²+(x+1)²=(x+2)²-2x+2;b) (2x-3) (2x+3)-1=5x+(x-2)²;

▪а) (x-1)²+(x+1)²=(x+2)²-2x+2;
x² - 2х + 1 + x² + 2х + 1 = x² + 4х + 4 - 2х + 2
2x² + 2 - x² - 2х - 6 = 0
x² - 2х - 4 = 0
D= b² - 4ac = 4 - 4 × (-4) = 4 + 16 = 20
$x1 = \frac{ - b + \sqrt{d} }{2a} = \frac{2 + \sqrt{20} }{2} = \frac{2 + 2 \sqrt{5} }{2} = \frac{2(1 + \sqrt{5)} }{2} = 1 + \sqrt{5}$
$x2 = \frac{ - b - \sqrt{d} }{2a} = \frac{2 - \sqrt{20} }{2} = \frac{2(1 - \sqrt{5)} }{2} = 1 - \sqrt{5}$

▪b) (2x-3) (2x+3)-1=5x+(x-2)²
4x² - 9 - 1 = 5x + x² - 4x + 4
4x² - x² - 10 - x - 4 = 0
3x² - x - 14 = 0
D= b² - 4ac = 1 - 4 × 3 × (-14) = 1 + 168 = 169
$x1 = \frac{ - b + \sqrt{d} }{2a} = \frac{1 + \sqrt{169} }{2 \times 3} = \frac{1 + 13}{6} = \frac{14}{6} = \frac{7}{3} = 2 \frac{1}{3}$
$x2 = \frac{1 - 13}{6} = \frac{ - 12}{6} = - 2$