ОДЗ :
0\\\\2(x-1,5)x(x+2)>0\\\\x(x-1,5)(x+2)>0" alt="\frac{2x-3}{x^{2}+2x }>0\\\\2(x-1,5)x(x+2)>0\\\\x(x-1,5)(x+2)>0" align="absmiddle" class="latex-formula">
- + - +
__________₀_________₀___________₀___________
- 2 0 1,5
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x ∈ (- 2 ; 0) ∪ (1,5 ; + ∞)
(\frac{1}{2})^{3}\\\\\frac{2x-3}{x^{2}+2x }>\frac{1}{8}\\\\\frac{2x-3}{x^{2}+2x }-\frac{1}{8}>0\\\\\frac{16x-24-x^{2}-2x }{8(x^{2}+2x) }>0\\\\\frac{-x^{2}+14x-24 }{8x(x+2)}>0\\\\\frac{x^{2}-14x+24 }{x(x+2)}<0\\\\\frac{(x-2)(x-12)}{x(x+2)}<0" alt="log_{\frac{1}{2}}\frac{2x-3}{x^{2}+2x }<3\\\\\frac{2x-3}{x^{2}+2x }>(\frac{1}{2})^{3}\\\\\frac{2x-3}{x^{2}+2x }>\frac{1}{8}\\\\\frac{2x-3}{x^{2}+2x }-\frac{1}{8}>0\\\\\frac{16x-24-x^{2}-2x }{8(x^{2}+2x) }>0\\\\\frac{-x^{2}+14x-24 }{8x(x+2)}>0\\\\\frac{x^{2}-14x+24 }{x(x+2)}<0\\\\\frac{(x-2)(x-12)}{x(x+2)}<0" align="absmiddle" class="latex-formula">
+ - + - +
_________₀_________₀_________₀_________₀_________
- 2 0 2 12
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x ∈ (- 2 ; 0) ∪ (2 ; 12)
Окончательный ответ : x ∈ (- 2 ; 0) ∪ (2 ; 12)