Question

Помогите решить пожалуйста

Answer 1

Ответ:

$1)\ \ \Delta =\left|\begin{array}{ccc}2&1&-1\\1&1&1\\3&-1&1\end{array}\right|=2\cdot (1+1)-(1-3)-(-1-3)=10\ne 0\\\\\\\Delta _{x}=\left|\begin{array}{ccc}1&1&-1\\6&1&1\\4&-1&1\end{array}\right|=1\cdot (1+1)-(6-4)-(-6-4)=10$

$\Delta _{y}=\left|\begin{array}{ccc}2&1&-1\\1&6&1\\3&4&1\end{array}\right|=2\cdot (6-4)-(1-3)-(4-18)=20\\\\\\\Delta _{z}=\left|\begin{array}{ccc}2&1&1\\1&1&6\\3&-1&4\end{array}\right|=2\cdot (4+6)-(4-18)+(-1-3)=30\\\\\\x=\dfrac{\Delta _{x}}{\Delta }=1\ ,\ \ y=\dfrac{\Delta _{y}}{\Delta }=2\ ,\ \ z=\dfrac{\Delta _{z}}{\Delta }=3$

$2)\ \ \displaystyle \frac{(2+3i)^2}{5-i}=\frac{4+12i+9i^2}{5-i}=\frac{(-5+12i)(5+i)}{(5-i)(5+i)}=\frac{-25-5i+60i+12i^2}{25-i^2}=\\\\\\=\frac{-37+55i}{26}=-\frac{37}{26}+\frac{55}{26}\, i$

$3)\ \ \vec{a}=(0;1;2)\ ,\ \ \vec{b}=(2;4;6)\\\\\vec{m}=3\vec{a}+\vec{b}=(0+2\, ;\, 3+4\, ;\, 6+6\, )=(2;7;12)\\\\|\vec{m}|=\sqrt{2^2+7^2+12^2}=\sqrt{197}$