Решение уравнений 1). (x-7)^2=(9-x). 2). (x+10)^2=(2-x)^2. 3).(x+9)^2=(10-x)^2

Решите задачу:

$1) {x}^{2} - 14x + 49 - 9 + x = 0 \\ {x}^{2} - 13x + 40 = 0 \\ d = 169 - 160 = 9 = 3 \\$

$x1 = \frac{14 + 3}{2} = \frac{17}{2} = 8 \frac{1}{2} = 8.5 \\ x2 = \frac{14 - 3}{2} = \frac{11}{2} = 5 \frac{1}{2} = 5.5$

$2) {x}^{2} + 20x + 100 = 4 - 4x + {x}^{2} \\ {x}^{2} + 20x + 100 - 4 + 4x - {x}^{2} = 0 \\ 24x + 96 = 0 \\ 24x = 0 - 96 = - 96 \\ x$

$x = - 96 \div 24 \\ x = - 4$

$3) {x}^{2} + 18x + 81 = 100 - 20x + {x}^{2} \\ {x}^{2} + 18x + 81 - 100 + 20x - {x}^{2} = 0 \\ 38x - 19 = 0 \\ 38x = 19 \\ x$

$x = 19 \div 38 = \frac{19}{38} = \frac{1}{2} = 0.5$