Решите это тоже, пожалуйста

Question

Решите это тоже, пожалуйста

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Answer 1

1 \\\\ \left \{ {{\frac{3+2x^2}{1+x^2}>0} \atop {\frac{5+4x^2}{1+x^2}>0}} \right.\\\\ \left \{ {{x \in (-\infty;\infty)} \atop { x \in (-\infty ;\infty)}} \right. \\\\ log_{a}(\frac{3+2x^2}{1+x^2}*\frac{5+4x^2}{1+x^2})>1\\\\ log_{a}(\frac{(3+2x^2)(5+4x^2)}{(1+x^2)^2})>1\\\\ \frac{(3+2x^2)(5+4x^2)}{(1+x^2)^2}>a\\\\ \frac{8x^4+22x^2+15}{x^4+2x^2+1}>a\\\\ 8x^4+22x^2+15>ax^4+2ax^2+a\\\\ " alt="log_{a}(\frac{3+2x^2}{1+x^2})+log_{a}(\frac{5+4x^2}{1+x^2})>1 \\\\ \left \{ {{\frac{3+2x^2}{1+x^2}>0} \atop {\frac{5+4x^2}{1+x^2}>0}} \right.\\\\ \left \{ {{x \in (-\infty;\infty)} \atop { x \in (-\infty ;\infty)}} \right. \\\\ log_{a}(\frac{3+2x^2}{1+x^2}*\frac{5+4x^2}{1+x^2})>1\\\\ log_{a}(\frac{(3+2x^2)(5+4x^2)}{(1+x^2)^2})>1\\\\ \frac{(3+2x^2)(5+4x^2)}{(1+x^2)^2}>a\\\\ \frac{8x^4+22x^2+15}{x^4+2x^2+1}>a\\\\ 8x^4+22x^2+15>ax^4+2ax^2+a\\\\ " align="absmiddle" class="latex-formula">
 0\\\\ D=(22-2a)^2-4(8-a)(15-a)=\sqrt{4(a-1)}=2\sqrt{a-1} \\\\ x^2=\frac{2a-22+2\sqrt{a-1}}{8-a} \\\\ x^2=\frac{2a-22-2\sqrt{a-1}}{8-a}\\\\ \left \{ {{\frac{2a-22+2\sqrt{a-1}}{8-a}<0} \atop {\frac{2a-22-2\sqrt{a-1}}{8-a}<0}} \right. \\\\ " alt="x^4(8-a)+x^2(22-2a)+15-a>0\\\\ D=(22-2a)^2-4(8-a)(15-a)=\sqrt{4(a-1)}=2\sqrt{a-1} \\\\ x^2=\frac{2a-22+2\sqrt{a-1}}{8-a} \\\\ x^2=\frac{2a-22-2\sqrt{a-1}}{8-a}\\\\ \left \{ {{\frac{2a-22+2\sqrt{a-1}}{8-a}<0} \atop {\frac{2a-22-2\sqrt{a-1}}{8-a}<0}} \right. \\\\ " align="absmiddle" class="latex-formula">
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