Question

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Answer 1

Объяснение:

7.

$\displaystyle\\1)\ (\sqrt[3]{\sqrt{x} }*x^{-\frac{3}{4} } )^{12}=(x^{\frac{1}{3}*\frac{1}{2}}*x^{-\frac{3}{4} })^{12}= (x^{\frac{1}{6}}*x^{-\frac{3}{4} })^{12}=(x^{\frac{1}{6}+(-\frac{3}{4})})^{12}=\\\\ =(x^\frac{2+(-9)}^{12}})^{12} =x^{\frac{-7*12}{12}}=x^{-7} .$

$2)\ (\sqrt[7]{m}*\sqrt[3]{m^{-2}} )^{-14}=(m^{\frac{1}{7}} *m^{-\frac{2}{3}})^{-14}=(m^{\frac{3+(-14)}^{21}}})^{-14}=m^{\frac{-11*(-14)}{21} } =m^{\frac{22}{3} }.$

8.

$cos\alpha =0,8\ \ \ \ \ \ \frac{3\pi }{2} < \alpha < 2\pi$

$\displaystyle\\1)\ sin\alpha =?\\\\sin^2\alpha +cos^2\alpha =1\\\\sin^2\alpha =1-cos^2\alpha =1-0,8^2=1-0,64=0,36.\\\\sin\alpha =б\sqrt{0,36}=б0,6.\\\\\frac{3\pi }{2} < \alpha < 2\pi \ \ \ \ \ \ \Rightarrow\\\\sin\alpha =-0,6.$

$\displaystyle\\2)\ tg(\frac{\pi }{4}+\alpha )=\frac{sin(\frac{\pi }{4}+\alpha ) }{cos(\frac{\pi }{4}+\alpha ) } =\frac{sin\frac{\pi }{4}*cos\alpha +cos\frac{\pi }{4}*sin\alpha }{cos\frac{\pi }{4}*cos\alpha -sin\frac{\pi }{4}*sin\alpha } =$

$\displaystyle\\=\frac{\frac{\sqrt{2} }{2}*cos\alpha +\frac{\sqrt{2} }{2} *sin\alpha }{\frac{\sqrt{2} }{2} *cos\alpha -\frac{\sqrt{2} }{2}*sin\alpha } =\frac{\frac{\sqrt{2} }{2}*(cos\alpha +sin\alpha ) }{\frac{\sqrt{2} }{2}*(cos\alpha -sin\alpha ) } =\frac{cos\alpha+sin\alpha }{cos\alpha -sin\alpha } =\\\\\\=\frac{0,8+(-0,6)}{0,8-(-0,6)}=\frac{0,2}{1,4} =\frac{1}{7} .$