Question

помогите с 9 заданием из егэшки

Answer 1

$25)\frac{\sqrt[9]{11}*11*\sqrt[18]{11}}{\sqrt[6]{11} }=\frac{11^{\frac{1}{9}}*11*11^{\frac{1}{18}} }{11^\frac{1}{6}}}=11^{\frac{1}{9}+1+\frac{1}{18}-\frac{1}{6}}=11^{\frac{18}{18}}=\boxed{11}\\\\\\26)\sqrt{108} *Cos^{2}\frac{\pi }{12}-\sqrt{27}=\sqrt{36*3}*\frac{1+Cos\frac{\pi }{6} }{2}-\sqrt{9*3} =\\\\=6\sqrt{3}*\frac{1}{2}*(1+\frac{\sqrt{3} }{2})-3\sqrt{3}=3\sqrt{3}*(1+\frac{\sqrt{3} }{2}-1)=\frac{3\sqrt{3}*\sqrt{3}}{2}=\\\\=\frac{9}{2} =\boxed{4,5}$

$27)7\sqrt{2} Sin\frac{15\pi }{8}*Cos\frac{15\pi }{8} =7\sqrt{2}*\frac{1}{2}*2 Sin\frac{15\pi }{8}*Cos\frac{15\pi }{8} =\\\\=\frac{7}{\sqrt{2} }*Sin\frac{15\pi }{4}=\frac{7}{\sqrt{2} } Sin(4\pi-\frac{\pi }{4})=-\frac{7}{\sqrt{2} }*Sin\frac{\pi }{4}=-\frac{7}{\sqrt{2}}*\frac{\sqrt{2} }{2}=\\\\=-\frac{7}{2}=\boxed{-3,5}$