Question

Помогите решить второе и четвертое задание

Answer 1

2)
$\frac{1 + i \sqrt{3} }{1 - i \sqrt{3} } = \frac{(1 + i \sqrt{3})(1 + i \sqrt{3} )}{(1 - i \sqrt{3})(1 + i \sqrt{3} ) } = \frac{ {(1 + i \sqrt{3} )}^{2} }{1 - 3 {i}^{2} } = \frac{1 + 2i \sqrt{3} + 3 {i}^{2} }{1 - 3 \times ( - 1)} = \frac{1 + 2i \sqrt{3} + 3 \times ( - 1)}{1 + 3} = \frac{ - 2 + 2i \sqrt{3} }{4} = \frac{ - 1 - i \sqrt{3} }{2}$
4)
y'(x) =
$= \frac{1}{ \sqrt{tg3x} } \times \frac{1}{2 \sqrt{tg3x} } \times \frac{1}{ { \cos}^{2}3x } \times 3 = \frac{3}{2} \times \frac{1}{tg3x} \times \frac{1}{ cos^{2} 3x}$
y'(pi/12) =
$= \frac{3}{2} \times \frac{1}{tg3x} \times \frac{1}{ cos^{2}3x } = \frac{3}{2} \times \frac{1}{tg(3 \times \frac{\pi}{12} )} \times \frac{1}{ \cos^{2} (3 \times \frac{\pi}{12}) } = \frac{3}{2} \times \frac{1}{tg \frac{\pi}{4} } \times \frac{1}{ \cos^{2} \frac{\pi}{4} } = \frac{3}{2} \times \frac{1}{1} \times \frac{1}{ \frac{1}{2} } = \frac{3}{2} \times 2 = 3$