0} \atop {2x^2-1\ne 1\; ,\; x+2>0}} \right. \; \left \{ {{(\sqrt2x-1)(\sqrt2x+1)>0} \atop {x^2\ne 1\; ,\; x>-2}} \right. \\\\\left \{ {{x\in (-\infty ,-1/\sqrt2)\cup (1/\sqrt2,+\infty )} \atop {x\in (-2,-1)\cup (-1,1)\cup (1,+\infty )}} \right. \; \to \\\\x\in (-2,-1)\cup (-1,-\frac{1}{\sqrt2})\cup (\frac{1}{\sqrt2},1)\cup (1,+\infty )\\\\x+2=\sqrt{2x^2-1}\\\\x^2+4x+4=2x^2-1\\\\x^2-4x-5=0\\\\D/4=4+5=9\; ,\; \; x_1=-1\notin ODZ\; ,\; x_2=5\\\\Otvet:\; \; x=5\; ." alt=" log\; _{2x^2-1}(x+2)=\frac{1}{2}\\\\ODZ:\; \; \left \{ {{2x^2-1>0} \atop {2x^2-1\ne 1\; ,\; x+2>0}} \right. \; \left \{ {{(\sqrt2x-1)(\sqrt2x+1)>0} \atop {x^2\ne 1\; ,\; x>-2}} \right. \\\\\left \{ {{x\in (-\infty ,-1/\sqrt2)\cup (1/\sqrt2,+\infty )} \atop {x\in (-2,-1)\cup (-1,1)\cup (1,+\infty )}} \right. \; \to \\\\x\in (-2,-1)\cup (-1,-\frac{1}{\sqrt2})\cup (\frac{1}{\sqrt2},1)\cup (1,+\infty )\\\\x+2=\sqrt{2x^2-1}\\\\x^2+4x+4=2x^2-1\\\\x^2-4x-5=0\\\\D/4=4+5=9\; ,\; \; x_1=-1\notin ODZ\; ,\; x_2=5\\\\Otvet:\; \; x=5\; ." align="absmiddle" class="latex-formula">