Question

Помогите решить СРООЧНО

Answer 1

$\sqrt{x} \times \sqrt{2 - x} = 2x \\ \sqrt{x \times (2 - x)} = 2x \\ \sqrt{2x - {x}^{2} } = 2x \\ ( \sqrt{2x - {x}^{2} } )^{2} = {(2x)}^{2} \\ 2x - {x}^{2} = {4x}^{2} \\ 2x - {x}^{2} - 4 {x}^{2} = 0 \\ 2x - {5x}^{2} = 0 \\ x(2 - 5x) = 0 \\ x = 0 \\ 2 - 5x = 0 \\ x = \frac{2}{5}$

проверка:

$\sqrt{0} \times \sqrt{2 - 0} = 2 \times 0 \\ 0 \sqrt{2} = 0 \\ 0 = 0$

$\sqrt{ \frac{2}{5} } \times \sqrt{2 - \frac{2}{5} } = 2 \times \frac{2}{5} \\ \sqrt{ \frac{2}{5} \times (2 - \frac{2}{5}) } = 0.8 \\ \sqrt{ \frac{2}{5} \times \frac{8}{5} } = 0.8 \\ \sqrt{ \frac{16}{25} } = 0.8 \\ \frac{4}{5} = 0.8 \\ 0.8 = 0.8$

ответ: x=0 , x=0,8