Question

помогите пожалуйста ​

Answer 1

Ответ:

1)

$\sqrt{x + 2} = 4$

${ \sqrt{x + 2} }^{2} = {4}^{2}$

$x + 2 = 16$

$x = 16 - 2$

$x = 14$

$\sqrt{14 + 2} = 4$

$\sqrt{16} = 4$

$4 = 4$

$x = 14$

2)

$\sqrt[3]{3 - {x}^{2} } = - 1$

${ \sqrt[3]{3 - {x}^{2} } }^{3} = {( - 1)}^{3}$

$3 - {x}^{2} = - {1}^{3}$

$3 - {x}^{2} = - 1$

$- {x}^{2} = - 1 - 3$

$- {x}^{2} = - 4$

${x}^{2} = 4$

$x = ± \sqrt{4}$

$x = ±2$

$x = - 2 \\ x = 2$

$x_{1} = - 2. x_{2} = 2$