Question

Помогите пожалуйста!!!!! ​

Answer 1

Ответ:

$log_2(3x+1)\cdot log_{0,5}(6x+2)\leq -6\ \ ,\ \ \ ODZ:\ \ 3x+1>0\ \ ,\ \ x>-\dfrac{1}{3}\\\\log_2(3x+1)\cdot \Big(-log_{2}(2\cdot(3x+1)\Big)\leq -6\\\\log_3(3x+1)\cdot (-1-log_2(3x+1))\leq -6\\\\-log^2_2(3x+1)-log_2(3x+1)\leq -6\\\\t=log_2(3x+1)\ \ ,\ \ \ t^2+t-6\leq 0\ \ ,\ \ \ t_1=-3\ ,\ \ t_2=2\\\\(t+3)(t-2)\leq 0\ \ ,\ \ \ \ -3\leq t\leq 2\ \ ,\ \-3\leq log_2(3x+1)\leq 2$

$\left\{\begin{array}{l}log_2(3x+1)\geq -3\\log_2(3x+1)\leq 2\end{array}\right\ \ \ \left\{\begin{array}{l}3x+1\geq 2^{-3}\\3x+1\leq 2^2\end{array}\right\ \ \ \left\{\begin{array}{l}3x+1\geq \dfrac{1}{8}\\3x+1\leq 4\end{array}\right\\\\\\\left\{\begin{array}{l}x\geq -\dfrac{7}{24}\\x\leq 1\end{array}\right\ \ \ \Rightarrow \ \ \ x\in \Big[\ -\dfrac{7}{24}\ ;\ 1\ \Big]\\\\\\a=-\dfrac{7}{24}\ \ ,\ \ b=1\\\\\\24a+b=24\cdot \Big(-\dfrac{7}{24}\Big)+1=-7+1=-6$