8*10.000: 15*5.625 = 0.94
По свойству логарифмов
, представим второе слагаемое неравенства следующим образом:
![\log_{0.5}(2(3x+1))=\log_{0.5}2+\log_{0.5}(3x+1)=\log_{0.5}(3x+1)-1](https://tex.z-dn.net/?f=%5Clog_%7B0.5%7D%282%283x%2B1%29%29%3D%5Clog_%7B0.5%7D2%2B%5Clog_%7B0.5%7D%283x%2B1%29%3D%5Clog_%7B0.5%7D%283x%2B1%29-1)
Мы получим
![\log_2(3x+1)\cdot \left[\log_{0.5}(3x+1)-1\right]<-6\\ \\ \log_{2}(3x+1)\cdot \log_{0.5}(3x+1)-\log_2(3x+1)+6<0](https://tex.z-dn.net/?f=%5Clog_2%283x%2B1%29%5Ccdot%20%5Cleft%5B%5Clog_%7B0.5%7D%283x%2B1%29-1%5Cright%5D%3C-6%5C%5C%20%5C%5C%20%5Clog_%7B2%7D%283x%2B1%29%5Ccdot%20%5Clog_%7B0.5%7D%283x%2B1%29-%5Clog_2%283x%2B1%29%2B6%3C0)
По свойству логарифмов
, тогда
![-\log_2^2(3x+1)-\log_2(3x+1)+6<0~~|\cdot (-1)\\ \\ \log_2^2(3x+1)+\log_2(3x+1)-6>0\\ \\ \left(\log_2(3x+1)+\dfrac{1}{2}\right)^2-\dfrac{25}{4}>0~~~\Rightarrow~~~ \left|\log_2(3x+1)+\dfrac{1}{2}\right|>\dfrac{5}{2}](https://tex.z-dn.net/?f=-%5Clog_2%5E2%283x%2B1%29-%5Clog_2%283x%2B1%29%2B6%3C0~~%7C%5Ccdot%20%28-1%29%5C%5C%20%5C%5C%20%5Clog_2%5E2%283x%2B1%29%2B%5Clog_2%283x%2B1%29-6%3E0%5C%5C%20%5C%5C%20%5Cleft%28%5Clog_2%283x%2B1%29%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E2-%5Cdfrac%7B25%7D%7B4%7D%3E0~~~%5CRightarrow~~~%20%5Cleft%7C%5Clog_2%283x%2B1%29%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%7C%3E%5Cdfrac%7B5%7D%7B2%7D)
Следующее неравенство эквивалентно совокупности неравенств
![\left[\begin{array}{ccc}\log_2(3x+1)+\dfrac{1}{2}>\dfrac{5}{2}\\ \\ \log_2(3x+1)+\dfrac{1}{2}<-\dfrac{5}{2}\end{array}\right~~~\Rightarrow~~~~\left[\begin{array}{ccc}\log_2(3x+1)>2\\ \\ \log_2(3x+1)<-3\end{array}\right~~\Rightarrow~\\\\ \\ \Rightarrow~~~\left[\begin{array}{ccc}3x+1>4\\ \\ 3x+1<\dfrac{1}{8}\end{array}\right~~~\Rightarrow~~~\left[\begin{array}{ccc}x>1\\ \\ x<-\dfrac{7}{24}\end{array}\right](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Clog_2%283x%2B1%29%2B%5Cdfrac%7B1%7D%7B2%7D%3E%5Cdfrac%7B5%7D%7B2%7D%5C%5C%20%5C%5C%20%5Clog_2%283x%2B1%29%2B%5Cdfrac%7B1%7D%7B2%7D%3C-%5Cdfrac%7B5%7D%7B2%7D%5Cend%7Barray%7D%5Cright~~~%5CRightarrow~~~~%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Clog_2%283x%2B1%29%3E2%5C%5C%20%5C%5C%20%5Clog_2%283x%2B1%29%3C-3%5Cend%7Barray%7D%5Cright~~%5CRightarrow~%5C%5C%5C%5C%20%5C%5C%20%5CRightarrow~~~%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3x%2B1%3E4%5C%5C%20%5C%5C%203x%2B1%3C%5Cdfrac%7B1%7D%7B8%7D%5Cend%7Barray%7D%5Cright~~~%5CRightarrow~~~%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%3E1%5C%5C%20%5C%5C%20x%3C-%5Cdfrac%7B7%7D%7B24%7D%5Cend%7Barray%7D%5Cright)
x ∈ (-∞; -7/24) U (1; +∞).
ОДЗ неравенства: ![\displaystyle \left \{ {{3x+1>0} \atop {6x+2>0}} \right. ~~~\Rightarrow~~~ 3x+1>0~~~\Rightarrow~~~ x>-\dfrac{1}{3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%20%5C%7B%20%7B%7B3x%2B1%3E0%7D%20%5Catop%20%7B6x%2B2%3E0%7D%7D%20%5Cright.%20~~~%5CRightarrow~~~%203x%2B1%3E0~~~%5CRightarrow~~~%20x%3E-%5Cdfrac%7B1%7D%7B3%7D)
С учетом ОДЗ, получаем ответ ![x \in \left(-\dfrac{1}{3};-\dfrac{7}{24}\right)\cup\left(1;+\infty\right).](https://tex.z-dn.net/?f=x%20%5Cin%20%5Cleft%28-%5Cdfrac%7B1%7D%7B3%7D%3B-%5Cdfrac%7B7%7D%7B24%7D%5Cright%29%5Ccup%5Cleft%281%3B%2B%5Cinfty%5Cright%29.)
<span>(3-m)(8+n)+(m-4)(n+6)
<u>24</u>+3n-8m<u>-mn+mn</u>+6m-4n<u>-24</u>
<u>3n</u>-8m+6m<u>-4n
</u>-n-2m
<u></u>Ответ: -n-2m
</span>