Распишите решение ПОДРОБНО номера 1. 2. 7 Спасибо

1)
$4 x + y = x + 2 \\ 2x - 2y = 3 - y \\ \\ 4x - x + y = 2 \\ 2x - 2y + y = 3 \\ \\ 3x + y= 2 \\ 2x - y = 3$
Складываем два уравнения:
$5x = 5 \\ y = 2x - 3 \\ \\ x = 1 \\ y = 2x - 3 \\ \\ x = 1 \\ y = 2 \times 1 - 3 = - 1$
Ответ: (1; -1)

2)
$f(2) = \frac{ {2}^{2} - 2 \times 2 - 3 }{2 - 3} = \frac{ - 3}{ - 1} = 3 \\ \\ f( - 1) = \frac{ {( - 1)}^{2} - 2 \times ( - 1) - 3}{ - 1 - 3} = \frac{1 + 2 - 3}{ - 4} = \frac{0}{ - 4} = 0 \\ \\ f(0) = \frac{ {0}^{2} - 2 \times 0 - 3}{0 - 3} = \frac{ - 3}{ - 3} = 1 \\ \\ f(2) - 3\times f( - 1) + 2f(0) = 3 - 3 \times 0 + 2 \times 1 = 3 + 2 = 5$
Ответ: 5.

7)
$2b(3 {a}^{2} + {b}^{2} ) = 6 {a}^{2} b + 2 {b}^{3} = {a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3} - {a}^{3} + 3 {a}^{2} b - 3a {b}^{2} + {b}^{3} = {(a + b)}^{3} - ( {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3} ) = {(a + b)}^{3} - {(a - b)}^{3}$