Question

Помогите пж с третьим заданием

Answer 1

$1) text{log}_{0,9}(x-5) geqslant text{log}_{0,9}11$

$0,9 < 1 Rightarrow y = text{log}_{0,9} downarrow \left { {bigg{x-5>0 } atop bigg{x-5 leqslant 11}} right. left { {bigg{x>5 } atop bigg{x leqslant 16}} right.$

$text{OTBET:} x in (5; 16]$

$2) log_{x}(x^{2} + 3x - 8) > 1\log_{x}(x^{2} + 3x - 8) > log_{x}x$

$D(y): x^{2} + 3x - 8 > 0; x = -1,5 + sqrt{41}; x in (-1,5 + sqrt{41}; + infty)$

$text{a)} x > 1 Rightarrow y = log_{x} uparrow\begin{equation*} begin{cases} x > 1, \ x^{2} + 3x - 8 > x. end{cases}end{equation*}$

$begin{equation*} begin{cases}x > 1, \ x_{1} = -4; x_{2} = 2. end{cases}end{equation*}$

$x in (2; + infty)$

При $x in (-1,5 + sqrt{41}; + infty)$, то $x in (2; + infty)$

$text{b)} x < 1 Rightarrow y = log_{x} downarrow \begin{equation*} begin{cases}x > 1, \ x^{2} + 3x - 8 < x. end{cases}end{equation*}$

$begin{equation*} begin{cases}0<x<1, \ x_{1} = -4; x_{2} = 2. end{cases}end{equation*}$

$x in (0; 1)$

При $x in (-1,5 + sqrt{41}; + infty)$, то $x notin (0; 1)$

$text{OTBET:} x in (2; + infty)$