Вычислите:а) б) в)

Решите задачу:

$-3 \sqrt{1,69} + \frac{1}{2} (\sqrt{6})^{2}=-3*1,3+\frac{1}{2}*6=-3,9+3=-0,9$

$-5 \sqrt{3 \frac{1}{16} }-2 \sqrt{3 \frac{6}{25} } = -5 \sqrt{\frac{49}{16} }-2 \sqrt{\frac{81}{25}}=\\\ =-5*\frac{7}{4}}-2*\frac{9}{5}}=-\frac{35}{4}}-\frac{18}{5}}=-8\frac{3}{4}}-3\frac{3}{5}}=\\\ =-8\frac{15}{20}}-3\frac{12}{20}}=-11\frac{27}{20}}=-12\frac{7}{20}}$

$(\sqrt{12} - \sqrt{2})^{2}=(\sqrt{12})^2-2\sqrt{12}*\sqrt{2}+(\sqrt{2})^2=12-2\sqrt{24}+2=14-4\sqrt{6}$