Question

срочно нужна помощь!!

Answer 1

Ответ:

1.

$x - y = 2 = > x = y + 2 \\ {3}^{ {x}^{2} + y} = \frac{1}{9} \\ \\ {3}^{ {x}^{2} + y } = {9}^{ - 1} \\ {3}^{ {x}^{2} + y } = {3}^{ - 2} \\ {x}^{2} + y = - 2 \\ ( {y + 2)}^{2} + y + 2 = 0 \\ {y}^{2} + 4y + 4 + y + 2 = 0 \\ {y}^{2} + 5y + 6 = 0$

$y1 = - 3 \\ y2 = - 2 \\ x1 = - 3 + 2 = - 1 \\ x2 = - 2 + 2 = 0 \\ ( - 1 ;- 3);(0 ;- 2)$

2.

$x + 2y = 3 = > x = 3 - 2y \\ {3}^{x - y} = 81 \\ {3}^{x - y} = {3}^{4} \\ x - y = 4 \\ 3 - 2y - y = 4 \\ - 3y = 1 \\ y = - \frac{1}{3} \\ x = 3 - 2 \times ( - \frac{1}{3} ) =3 \frac{2}{3} \\ (3 \frac{2}{3}; - \frac{1}{3} )$

3.

$x + y = 1 \: ; \: \: \: \: \: \: \: {2}^{x - y} = 8 \\ {2}^{x - y} = {2}^{3} \\ x - y = 3 \\ x = y + 3 \\ y + 3 + y = 1 \\ 2y = - 2 \\ y = - 1 \\ x = - 1 + 3 = 2 \\ (2; - 1)$