0 \: \: x > - 2 \\ x + 3 > 0 \: \: \: x > - 3 \\ log_{0.5}((x + 2) (x + 3)) = log_{0.5}(3) - log_{ 0.5 }(0.5) \\ log_{0.5}((x + 2)(x + 3)) = log_{0.5}( \frac{3}{0.5} ) \\ (x + 2)(x + 3) = 6 \\ {x}^{2} + 3x + 2x + 6 = 6 \\ {x}^{2} + 5x = 0 \\ x(x + 5) = 0 \\ x = 0 \\ x = - 5" alt=" log_{0.5}(x + 2) + log_{0.5}(x + 3) = log_{0.5}(3) - 1 \\ odz \\ x + 2 > 0 \: \: x > - 2 \\ x + 3 > 0 \: \: \: x > - 3 \\ log_{0.5}((x + 2) (x + 3)) = log_{0.5}(3) - log_{ 0.5 }(0.5) \\ log_{0.5}((x + 2)(x + 3)) = log_{0.5}( \frac{3}{0.5} ) \\ (x + 2)(x + 3) = 6 \\ {x}^{2} + 3x + 2x + 6 = 6 \\ {x}^{2} + 5x = 0 \\ x(x + 5) = 0 \\ x = 0 \\ x = - 5" align="absmiddle" class="latex-formula">
х=-5 посторонний корень ,т.к не принадлежит ОДЗ
Ответ : х=0