8-(х-3)(х+3)=10-(х-1)² ,(2х+1)²-(2х-3)²=4(7х-5)

Решите задачу:

$8 - (x - 3)( x + 3) = 10 - (x - 1 {)}^{2} \\ 8 - {x}^{2} + 9 = 10 - {x}^{2} + 2x - 1 \\ 8 + 9 = 10 + 2x - 1 \\ 17 = 9 + {2}x \\ - 2x = 9 - 17 \\ - 2x = - 8 \\ x = 4$

$(2x + 1 {)}^{2} - (2x - 3 {)}^{2} = 4(7x - 5 )\\ 4 {x}^{2} + 4x + 1 - (4 {x}^{2} - 12x + 9) = 28x - 20 \\ 4 {x}^{2} + 4x + 1 - 4 {x}^{2} + 12x - 9 = 28x - 20 \\ 16x - 8 = 28x - 20 \\ 16x - 28x = - 20 + 8 \\ - 12x = - 12 \\ x = 1$