Question

Решите очень надо кр даю 80 б. ​

Answer 1

1.

а

$\frac{3 \times \sqrt[3]{ \frac{8}{27} } }{2.5} + \frac{ \sqrt{0.25} }{2.5} = \frac{3 \times \frac{2}{3} }{2.5} + \frac{0.5}{2.5} = \\ = \frac{2}{2.5} + \frac{0.5}{2.5} = \frac{2.5}{2.5} = 1$

б

$1.4 {a}^{ \frac{1}{7} } \div 2 {a}^{ \frac{8}{7} } = 0.7 {a}^{ \frac{1}{7} - \frac{8}{7} } = 0.7 { a }^{ - 1} = \frac{7}{10a} \\$

в

${2}^{ log_{2}(7) } \times log_{3}( \frac{1}{9} ) = 7 \times ( - 2) = - 14 \\$

г

$log_{2}(10) - 2 log_{2}(5) + log_{2}(40) = log_{2}( \frac{10 \times 40}{ {5}^{2} } ) = \\ = log_{2}(16) = 4$

2.

$\sin( \alpha ) = 0.8 \\ \\ \cos( \alpha ) < 0 \\ \cos( \alpha ) = \sqrt{1 - \sin {}^{2} ( \alpha ) } \\ \cos( \alpha ) = - \sqrt{1 - 0.64} = - \sqrt{0.36} = - 0.6$

3.

$\cos {}^{2} (15^{\circ} ) - \sin {}^{2} (15^{\circ} ) = \ \\ = cos(2 \times 15^{\circ} ) = \cos(30^{\circ} ) = \frac{1}{2} = 0.5$

4.

а

$( \frac{1}{125} ) {}^{0.2x + 1} = 25 \\ {5}^{ - 3(0.2x + 1)} = {5}^{2} \\ - 3(0.2x + 1) = 2 \\ - 0.6x - 3 = 2 \\ - 0.6x = 5 \\ - \frac{3x}{5} = 5 \\ x = - \frac{25}{3}$

б

$log_{2}(2x - 4) = 7 \\ 2x - 4 = {2}^{7} \\ 2x - 4 = 128 \\ 2x = 132 \\ x = 66$

в

$log_{ \frac{1}{7} }(2x + 5) - log_{ \frac{1}{7} }(6) = log_{ \frac{1}{7} }(2) \\ log_{ \frac{1}{7} }( \frac{2x + 5}{6} ) = log_{ \frac{1}{7} }(2) \\ \frac{2x + 5}{6} = 2 \\ 2x + 5 = 12 \\ 2x = 7 \\ x = 3.5$

г

$\sqrt{ {x}^{2} - 6 } = \sqrt{ - 5x} \\ {x}^{2} - 6 = - 5x \\ {x}^{2} + 5x - 6 = 0 \\ D = 25 + 24 = 49 \\ x_1 \frac{ - 5 + 7}{2} = 1 \\ x_2 = - 6$

Проверка:

$x_1 = 1 \\ \sqrt{1 - 6} = \sqrt{ - 5}$

по знаком корня отрицательные числа, не подходит

$x_2 = - 6 \\ \sqrt{36 - 6} = \sqrt{30} \\ \sqrt{30} = \sqrt{30}$

ответ: -6

д

$2 \sin(x) + 1 = 0 \\ \sin(x) = - \frac{1}{2} \\ x_1 = - \frac{\pi}{6} + 2 \pi \: n \\ x_2 = - \frac{5\pi}{6} + 2\pi \: n \\ n\in \: Z$

Эти же корни в градусах:

$x_1 = 330^{\circ} + 360^{\circ} n \\ x_2 = 210^{\circ} + 360^{\circ} n$

ответ: 210°

5.

а

$log_{ \frac{1}{2} }(2x + 5) - 3 \\ \\ \\ 2x + 5 0 \\ x - 2.5 \\ \\ \\ \frac{1}{2} < 1$

знак меняется

$2x + 5 < 8 \\ 2x < 3 \\ x < 1.5$

Пересекаем с ОДЗ

$x\in( - 2.5;1.5)$

б

$( \frac{1}{4} ) {}^{x} - {2}^{1 - x} - 8 < 0 \\ {2}^{ - 2x} - 2 \times {2}^{ - x} - 8 < 0 \\ \\ {2}^{ - x} = t \\ \\ t {}^{2} - 2t - 8 < 0 \\ D= 4 + 32 = 36 \\ t_1 = \frac{2 + 6}{2} = 4 \\ t_2 = - 2 \\ + \: \: \: \: \: \: \: - \: \: \: \: \: \: + \\ - -( - 2) - -4 - - \\ - 2 < t < 4 \\ \\ 1)t - 2 \\ {2}^{ - x} - 2 \\ x\in \: R\\ \\ 2)t < 4 \\ {2}^{ - x} < 4 \\ - x < 2 \\ x - 2$

$x\in( - 2 ;+ \infty )$

в

$\frac{ {x}^{2} + 2x - 3}{(x - 7)(x + 5)} < 0 \\ \\ \\ {x}^{2} + 2x - 3 = 0 \\ d = 4 + 12 = 16 \\ x1 = \frac{ - 2 + 4}{2} = 1 \\ x2 = - 3 \\ \\ \\ \frac{(x - 1)(x + 3)}{(x - 7)(x + 5)} < 0 \\ + \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: \: \: \: + \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: + \\ - - ( - 5) -( - 3) - - 1 - -7 - - \\ x\in( - 5; - 3)U(1;7)$