Помогите расписать, пожалуйста :с

$\frac{x^2+6x}{x-2} \leq 0 \\ \left \{ {{x(x+6) \leq 0} \atop {x-2\ \textless \ 0}} \right. \\ \left \{ {{x_1 \leq 0, x_2 \leq -6} \atop {x\ \textless \ 2}} \right.$
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$\frac{x(x+1)}{x-9} \ \textgreater \ 0 \\ \left \{ {{x(x+1) \geq 0} \atop {x-9\ \textgreater \ 0}} \right. \\ \left \{ {{x_1 \geq 0, x_2 \geq -1} \atop {x\ \textgreater \ 9}} \right.$
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$\frac{3x-2}{2x-3}\ \textgreater \ 3 \\ \frac{3x-2}{2x-3}-3\ \textgreater \ 0 \\ \frac{3x-2-3*(2x-3)}{2x-3}\ \textgreater \ 0 \\ \left \{ {{3x-2-6x+9 \geq 0} \atop {2x-3\ \textgreater \ 0}} \right. \\ \left \{ {{-6x \geq -7} \atop {2x-3\ \textgreater \ 0}} \right. \\ \left \{ {{x \geq 1 \frac{1}{6} } \atop {x\ \textgreater \ 1,5}} \right.$
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$\frac{x+3}{x-2}\ \textless \ 1 \\ \frac{x+3-x+2}{x-2} \ \textless \ 0 \\ \left \{ {{0x\ \textless \ 5} \atop {x\ \textless \ 2}} \right.$
Решений не имеет.
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$x^2+6x+9 \geq 0 \\ (x+3)^2 \geq 0 \\ x \geq 3, x \geq -3$
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$-4x^2+20x\ \textgreater \ 25 \\ -4x^2+20x-25\ \textgreater \ 0 \\ 4x^2-20x+25\ \textgreater \ 0 \\ (2x-5)^2\ \textgreater \ 0 \\ x\ \textgreater \ 2,5$