Question

помогите решить пожалуйста

Answer 1

Аапаааппппррраппвпаааппвраапаарпппп

Answer 2

$2)\; \; log_{\pi }(arccos(-\frac{1}{2})-arctg(-\sqrt3))=log_{\pi }(\pi -arccos\frac{1}{2}+arctg\sqrt3)=\\\\=\log_{\pi }(\pi -\frac{\pi}{3}+\frac{\pi }{3})=log_{\pi }\pi =1\\\\3)\; \; 15\, log_6\left (31-\sqrt[3]{3\sqrt5-5\sqrt2}\cdot \sqrt[6]{95+30\sqrt{10}}\cdot \sqrt[6]{625}\right )=\\\\=15\, log_6\left (31-\sqrt[6]{(3\sqrt5-5\sqrt2)^2\cdot (95+30\sqrt{10})}\cdot \sqrt[6]{5^4}\right )=\\\\=15\, log_6\left (31-\sqrt[6]{(95-30\sqrt{10})(95+30\sqrt{10})}\cdot \sqrt[6]{5^4}\right )=\\\\=15\, \left (31-\sqrt[6]{95^2-900\cdot 10}\cdot \sqrt[6]{5^4}\right )=$

$=15\, log_6\left (31-\sqrt[6]{25}\cdot \sqrt[6]{5^4}\right )=15\, log_6\left (31-\sqrt[6]{5^6}\right )=\\\\=15\, log_6(31-5)=15\, log_626\; .\\\\4)\; \; \frac{log_5400}{log\, _{0,64}\, 5}-\frac{log_580}{log\, _{3,2}\, 5}+log_525=\\\\\star \; \; log_5400=log_5(25\cdot 16)=log_5(5^2\cdot 2^4)=log_55^2+log2^4=\\=2log_55+4log_52=2+4log_52\\\\\star \; \; log\, _{0,64}\, 5=\frac{1}{log_50,64}=[0,64=\frac{64}{100}=\frac{16}{25}=\left (\frac{4}{5}\right )^2=\frac{2^4}{5^2}\; ]=$

$=\frac{1}{log_5\frac{2^4}{5^2}}=\frac{1}{log_52^4-log_55^2}=\frac{1}{4log_52-2}\; ;\\\\\star \; \; log_580=log_5(16\cdot 5)=log_5(2^4\cdot 5)=4log_52+1\; ;\\\\\star \; \; log\, _{3,2}\, 5=\frac{1}{log_53,2}=[\; 3,2=\frac{32}{10}=\frac{2^5}{2\cdot 5}=\frac{2^4}{5}\; ]=\frac{1}{log_5\frac{2^4}{5}}=\frac{1}{4log_52-1}\; ;\\\\\star \; \; \log_525=log_55^2=2log_55=2\; .\\------------------------------\\\\=(2+4log_52)\cdot (4log_52-2)-(4log_52+1)(4log_52-1)+2=\\\\=[t=log_52]=(2+4t)(4t-2)-(4t+1)(4t-1)+2=\\\\=(16t^2-4)-(16t^2-1)+2=\\\\=16t^2-4-16t^2+2+1=-1$