Question

Решить неравенства a) b) c) d) e) возвести в степень a) b) вычислить a) b) выражение a) sin(n+t) b) cos() c) tg ( d) ctg ( найти основной периуд функции y=cos 3x найти значение функций при x= y=cosx при x=

Answer 1

$a) \ \ 2^x\ \textless \ \frac{1}2\\ 2^x\ \textless \ 2^{-1}\\ x\ \textless \ -1\\ \\ b) \ \ 3^{2x-1}-3^{2x-3}\ \textless \ \frac{8}3\\ 3^{2x}*3^{-1}-3^{2x}*3^{-3}\ \textless \ \frac{8}3\\ t=3^{2x}\\ \frac{1}{3}t-\frac{1}{27}t\ \textless \ \frac{8}{3}\\ \\ \frac{8}{27}t\ \textless \ \frac{8}{3}\\ t\ \textless \ 9\\ 3^{2x}\ \textless \ 9\\ 3^{2x}\ \textless \ 3^2\\\\ 2x\ \textless \ 2\\ x\ \textless \ 1\\$
$c) \ \ 17^{\frac{x}{y-8}}\geq17\\ \frac{x}{y-8}\geq1\\ x\geq y-8\\ \\ d) log_{0,2}x\ \textless \ 3\\ \frac{log_3x}{log_30,2}\ \textless \ log_327\\ log_3x\ \textless \ log_327*log_30,2\\ log_3x\ \textless \ log_3(27*log_30,2)\\ x\ \textless \ 27log_30,2\\ \\ e) \ \ log_{0,3}(x^2+22)\ \textless \ log_{0,2}13x\\ x^2+22\ \textgreater \ 13x\\ x^2-13x+22\ \textgreater \ 0\\ x^2-13x+22=0\\ D=(-13)^2-4*1*22=81=9^2\\ x_1=\frac{13+9}{2}=11; \ \ \ x_2=\frac{13-9}{2}=2\\ x\in(-\infty;2)\cup(11;\infty)$

$a) \ \ (\frac{18a^9l^3c^8}{3x^4b^4c^4})^{-3}=\\ \\ =\frac{18^{-3}a^{9*(-3)}l^{3*(-3)}c^{4*(-3)}}{3^{-3}x^{4*(-3)}b^{4*(-3)}}=\\ \\ \frac{18^{-3}a^{-27}l^{-9}c^{-12}}{3^{-3}x^{-12}b^{-12}}=\\ =\frac{27(xb)^{12}}{18^3(a^9l^3c^4)^3}\\ \\ b) \ \ (\frac{7a^{15}b^{19}c^{-24}}{21a^1b^{-2}c^4})^{\frac{2}7}=\\ \\ =(\frac{a^{14}b^{21}c^{-28}}{3})^{\frac{2}7}=3^{-\frac{2}7}a^4b^6c^{-8}\\$

$a) \ \ \sqrt{x81}*\sqrt{4}+\sqrt{4}\sqrt[3]{343}=9\sqrt{x}*2+2*7=14+18\sqrt{x}\\ \\ b) \ \ \sqrt{9+6*1+1}=\sqrt{16}=4\\ \\ \\ b)\ \ cos(90^0+t)=-sin \ t \\\\ c) \ \ tg(\frac{3}2\pi-t)=ctg \ t\\ d) \ \ ctg (360^0-t)=-ctg \ t\\$

y=cos(3x)
Период функции y=cos(x) T=2π
cos(x)=cos(x+T)
cos(3x)=cos(3(x+K))
3(x+K)=2π
3x+3K=2π
x=2/3π

$y=2sin(x-\frac{\pi}{6})+1; \ \ \ x=\frac{4\pi}3\\ y=2sin(\frac{4\pi}3-\frac{\pi}{6})+1=2sin(\frac{14\pi}{12})+1=2sin(\frac{\pi}{6})+1=2*0,5+1=2\\ \\ \\ y=cos(x^2); \ \ \ x=\pi\\ y=cos(\pi^2)=-0,903$