Question

решите уравнение 7 класс
(3x-1)^2-8(x+1)^2=(x+2)(x-2)

Answer 1

Решение задания прилагаю

Answer 2

$\boxed{ {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} } \\ \boxed{ {(a + b) }^{2} = {a}^{2} + 2ab + {b}^{2} } \\ \boxed{ {a}^{2} - {b}^{2} = (a + b)(a - b)}$

${(3x - 1)}^{2} - 8{(x + 1)}^{2} = (x + 2)(x - 2) \\ 9 {x}^{2} - 6x + 1 - 8( {x}^{2} + 2x + 1) = {x}^{2} - 4 \\ 9 {x}^{2} - 6x + 1 - 8 {x}^{2} - 16x - 8 = {x}^{2} - 4 \\ 9 {x}^{2} - 8 {x}^{2} - {x}^{2} - 6x - 16x = - 4 - 1 + 8 \\ - 22x = 3 \\ \boldsymbol{x = - \frac{3}{22} }$