Question

!Допоможіть!, будь ласка.Тести на картинкі)

Answer 1

1.

$\sqrt[3]{8} + {27}^{ \frac{1}{3} } = \sqrt[3]{ {2}^{3} } + ({3}^{3} ) {}^{ \frac{1}{3} } = 2 + 3 = 5$

Ответ: В

2.

х - угол 4 четверти

$\cos(x) = \frac{5}{13} \\ \cos {}^{2} (x) + \sin {}^{2} (x) = 1 \: \: | \div \cos {}^{2} (x) \\ \frac{ \cos {}^{2} (x) }{ \cos {}^{2} (x) } + \frac{ \sin {}^{2} (x) }{ \cos {}^{2} (x) } = \frac{1}{ \cos {}^{2} (x) } \\ 1 + \tan {}^{2} (x) = \frac{1}{ \cos {}^{2} (x) } \\ \tan {}^{2} (x) = \frac{1}{ \cos {}^{2} (x) } - 1 \\ \tan(x) = - \sqrt{ \frac{1}{ \cos {}^{2} (x) } - 1} = - \sqrt{ \frac{1}{( \frac{5}{13} ) {}^{2} } - 1 } = - \sqrt{ \frac{1}{ \frac{25}{169} } - 1 } = \\ = - \sqrt{ \frac{169}{25} - 1} = - \sqrt{ \frac{169 - 25}{25} } = - \sqrt{ \frac{144}{25} } = - \sqrt{ \frac{ {12}^{2} }{ {5}^{2} } } = - \frac{12}{5} = - 2.4$

Ответ: Г

3.

$2 \sin(x) = 1 \\ \sin(x) = \frac{1}{2} \\ x = ( - 1) {}^{n} arcsin( \frac{1}{2} ) + \pi n \: , \: n \: \epsilon \: z\\ x = ( - 1) {}^{n} \frac{\pi}{6} + \pi n \: , \: n \: \epsilon \: z$

Ответ: Г