\(\dfrac{{{x^2} + 2x-3}}{{x{}^2 + 2x}} < 0.\)
Решим неравенство методом интервалов. Найдём нули числителя:
\({x^2} + 2x-3 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,}\\{x = -3.}\end{array}} \right.\)
Найдём нули знаменателя:
\({x^2} + 2x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\left( {x + 2} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,\,\,\,\,\,\,\,}\\{x + 2 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,\,\,\,\,}\\{x = -2.}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-3;-2} \right) \cup \left( {0;1} \right).\)
Ответ: \(\left( {-3;\;-2} \right) \cup \left( {0;\;1} \right).\)