Решите уравнение и неравенство пожалуйста

Question 1

Question 2

плохо видно основание логарифма

Question 3

основание log x

Question 4

Решите задачу:

$\sqrt{x^2-2x-4}=\sqrt{2x^2-6x-1}, \\ \left \{ {{x^2-2x-4\geq0,} \atop {2x^2-6x-1\geq0;}} \right. \left \{ {{D=5, \left [ {{x\leq1-\sqrt{5},} \atop {x\geq1+\sqrt{5};}} \right. } \atop {D=11, \left [ {{x\leq\frac{3-\sqrt{11}}{2},} \atop {x\geq\frac{3+\sqrt{11}}{2}}} \right. }} \right. \left [ {{x\leq1-\sqrt{5},} \atop {x\geq1+\sqrt{5};}} \right. \\ x^2-2x-4=2x^2-6x-1, \\ x^2-4x+3=0, \\ x_1=1, x_2=3, \\ 1-\sqrt{5}\ \textless \ 1\ \textless \ 1+\sqrt{5}, \\ 1-\sqrt{5}\ \textless \ 3\ \textless \ 1+\sqrt{5}, \\ x\in\varnothing.$

$\log_x\frac{2x+5}{4(x-1)}\leq0, \\ \left \{ {{\frac{2x+5}{4(x-1)}\ \textgreater \ 0,} \atop {4(x-1) \neq 0;}} \right. \\ (2x+5)(x-1)\ \textgreater \ 0, \\ \left [ {{x\ \textless \ -2,5,} \atop {x\ \textgreater \ 1;}} \right.$
$\left [ {{ \left \{ {{0\ \textless \ x\ \textless \ 1,} \atop {\frac{2x+5}{4(x-1)} \geq 1,}} \right. } \atop {\left \{ {{x\ \textgreater \ 1,} \atop {\frac{2x+5}{4(x-1)}\leq1,}} \right.}} \right. \left [ {{ \left \{ {{0\ \textless \ x\ \textless \ 1,} \atop {\frac{2x+5}{4(x-1)}-1\geq0,}} \right. } \atop {\left \{ {{x\ \textgreater \ 1,} \atop {\frac{2x+5}{4(x-1)}-1\leq0,}} \right.}} \right. \left [ {{ \left \{ {{0\ \textless \ x\ \textless \ 1,} \atop {\frac{-2x+9}{4(x-1)}\geq0,}} \right. } \atop {\left \{ {{x\ \textgreater \ 1,} \atop {\frac{-2x+9}{4(x-1)}\leq0,}} \right.}} \right.$
$\left [ {{ \left \{ {{0\ \textless \ x\ \textless \ 1,} \atop {(2x-9)(x-1)\leq0,}} \right. } \atop {\left \{ {{x\ \textgreater \ 1,} \atop {(2x-9)(x-1)\geq0,}} \right.}} \right. \left [ {{ \left \{ {{0\ \textless \ x\ \textless \ 1,} \atop {1\ \textless \ x\leq 4,5,}} \right. } \atop {\left \{ {{x\ \textgreater \ 1,} \atop { \left [ {{x\ \textless \ 1,} \atop {x \geq 4,5;}} \right. }} \right.}} \right. \left [ {{x\in\varnothing,} \atop {x \geq 4,5; }} \right.}} \right. \\ x \geq 4,5.$