Sin² 2x/3=3/4
(1-cos4x/3)/2=3/4
1-cos 4x/3=3/2
cos 4x/3=-1/2
4x/3=+-(π/3)+2πn
x=+-(π/4)+3/2πn
Х2+6х+8
х2+4х+2х+8
х*(х+4)+2*(х+4)
(х+2) * (х+4)
m3-n
![\displaystyle\mathtt{\frac{3-0,25^x}{2-2^{-x}}\geq1,5;~\left\{{{3-0,25^x\geq3-1,5*2^{-x}}\atop{2-2^{-x}\ \textgreater \ 0}}\right\left\{{{(\frac{1}{4})^x-1,5*(\frac{1}{2})^x\leq0}\atop{x\ \textgreater \ -1}}\right}\\\mathtt{\left\{{{(\frac{1}{2})^x[(\frac{1}{2})^x-1,5]\leq0}\atop{x\ \textgreater \ -1}}\right\left\{{{(\frac{1}{2})^x-(\frac{1}{2})^{\log_{\frac{1}{2}}1,5}\leq0}\atop{x\ \textgreater \ -1}}\right\left\{{{x\leq\log_2\frac{2}{3}}\atop{x\ \textgreater \ -1}}\right}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7B%5Cfrac%7B3-0%2C25%5Ex%7D%7B2-2%5E%7B-x%7D%7D%5Cgeq1%2C5%3B~%5Cleft%5C%7B%7B%7B3-0%2C25%5Ex%5Cgeq3-1%2C5%2A2%5E%7B-x%7D%7D%5Catop%7B2-2%5E%7B-x%7D%5C+%5Ctextgreater+%5C+0%7D%7D%5Cright%5Cleft%5C%7B%7B%7B%28%5Cfrac%7B1%7D%7B4%7D%29%5Ex-1%2C5%2A%28%5Cfrac%7B1%7D%7B2%7D%29%5Ex%5Cleq0%7D%5Catop%7Bx%5C+%5Ctextgreater+%5C+-1%7D%7D%5Cright%7D%5C%5C%5Cmathtt%7B%5Cleft%5C%7B%7B%7B%28%5Cfrac%7B1%7D%7B2%7D%29%5Ex%5B%28%5Cfrac%7B1%7D%7B2%7D%29%5Ex-1%2C5%5D%5Cleq0%7D%5Catop%7Bx%5C+%5Ctextgreater+%5C+-1%7D%7D%5Cright%5Cleft%5C%7B%7B%7B%28%5Cfrac%7B1%7D%7B2%7D%29%5Ex-%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%5Clog_%7B%5Cfrac%7B1%7D%7B2%7D%7D1%2C5%7D%5Cleq0%7D%5Catop%7Bx%5C+%5Ctextgreater+%5C+-1%7D%7D%5Cright%5Cleft%5C%7B%7B%7Bx%5Cleq%5Clog_2%5Cfrac%7B2%7D%7B3%7D%7D%5Catop%7Bx%5C+%5Ctextgreater+%5C+-1%7D%7D%5Cright%7D)
сравним логарифм с –1, чтобы выяснить, как они расположены на числовой прямой относительно друг друга
![\mathtt{-1~V~\log_2\frac{2}{3};~\log_2\frac{1}{2}~V~\log_2\frac{2}{3};~\frac{1}{2}~V~\frac{2}{3};~3~V~4;~3\ \textless \ 4}](https://tex.z-dn.net/?f=%5Cmathtt%7B-1~V~%5Clog_2%5Cfrac%7B2%7D%7B3%7D%3B~%5Clog_2%5Cfrac%7B1%7D%7B2%7D~V~%5Clog_2%5Cfrac%7B2%7D%7B3%7D%3B~%5Cfrac%7B1%7D%7B2%7D~V~%5Cfrac%7B2%7D%7B3%7D%3B~3~V~4%3B~3%5C+%5Ctextless+%5C+4%7D)
следовательно, логарифм лежит правее
ОТВЕТ:
![\mathtt{x\in(-1;1-\log_23]}](https://tex.z-dn.net/?f=%5Cmathtt%7Bx%5Cin%28-1%3B1-%5Clog_23%5D%7D)
A1=(5-x)/2
a2=(4x-1)/2
a3=4x
d=a2-a1=a3-a2
(4x-1)/2-(5-x)/2=4x-(4x-1)/2
4x-1-5+x=8x-4x+1
5x-4x=1+6
x=7
a2=(4*7-1)/2=(28-1)/2=27/2=13,5
Перенесем с противоположным знаком и будет 0, 4а=-4 и а =-10