Question

111111111111111111111111111111111111111111111111111111111

Answer 1

ОДЗ:
$2x-2 \ \textgreater \ 0 \ \Rightarrow \ 2x\ \textgreater \ 2 \ \Rightarrow \ \boxed{x\ \textgreater \ 1} \\ \\ 8x^2 -23x+15 \neq 1 \ \Rightarrow \ 8x^2 -23+14 \neq 0; \\ x_{1,2} = \frac{23 \pm \sqrt{23^2 -4 \cdot 8 \cdot 14}}{2 \cdot 8}=\frac{23 \pm \sqrt{529 - 448}}{16}=\frac{23 \pm 9}{16}; \ \ \boxed{x \neq 2}, \ \ x \neq \frac{7}{8} \ \textless \ 1 \\ \\ 8x^2 -23x +15 \ \textgreater \ 0; \ \ 8x^2 -23x+15=0; \ \ x_{3,4}=\frac{23 \pm \sqrt{529-480}}{16}=\frac{23 \pm 7}{16}; \\ x_3=\frac{15}{8}, \ x_4=1 \ \Rightarrow \ x\ \textless \ 1 ; \ \ \ \boxed{x\ \textgreater \ \frac{15}{8}}$

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             1                  15/8

$\frac{15}{8} \ \textless \ x \ \textless \ 2, \ \ x\ \textgreater \ 2$

$1) 8x^2 -23x+15 \ \textgreater \ 1 \ \Rightarrow \ x\ \textgreater \ 2 \\ \log_{8x^2 -23x+15} (2x-2) \leq 0 \\\\ \log_{8x^2 -23x+15} (2x-2)\leq \log_{8x^2-23x+15}1 \\ \\ 2x-2 \leq 1 \ \Rightarrow \ 2x \leq 3 \Rightarrow \ x \leq \frac{3}{2}$

Не удовлетворяет ОДЗ


$2) \ 0\ \textless \ 8x^2 -23x+15\ \textless \ 1 \\ \\ \left \{ {{x\ \textgreater \ \frac{15}{8}} \atop {\frac{7}{8} \ \textless \ x\ \textless \ 2}} \right. \ \Rightarrow \ \frac{15}{8}\ \textless \ x\ \textless \ 2 \\ \\ \log_{8x^2 -23x+15} (2x-2) \leq \log_{8x^2-23x+15}1 \\ \\ 2x-2 \geq 1 \\ \\ x \geq \frac{3}{2}$


Ответом будет пересечение $\frac{15}{8}\ \textless \ x\ \textless \ 2 \ \cup\ x \geq \frac{3}{2}$

Сравним  $\frac{15}{8} \ \textgreater \ \frac{3}{2} \ \ \ (\frac{15}{8} \ \textgreater \ \frac{12}{8})$


ОТВЕТ: $\frac{15}{8} \ \textless \ x \ \textless \ 2 \\ \\ (\frac{15}{8} ; 2)$