[tex]\displaystyle\bf\\2^{2x+1}-3^{2x+1} > 3^{2x} -7\cdot 2^{2x} \\\\\\2^{2x+1}+7\cdot 2^{2x} > 3^{2x+1} + 3^{2x} \\\\\\2^{2x} \cdot(2+7) > 3^{2x} \cdot(3+1)\\\\\\2^{2x} \cdot 9 > 3^{2x} \cdot 4 \ |: \ 2^{2x} > 0 \\\\\\\frac{2^{2x} \cdot 9}{2^{2x} } > \frac{3^{2x} \cdot 4}{2^{2x} } \\\\\\\Big(\frac{3}{2} \Big)^{2x} \cdot 4 < 9\\\\\\\Big(\frac{3}{2} \Big)^{2x} < \frac{9}{4} \\\\\\\Big(\frac{3}{2} \Big)^{2x} < \Big(\frac{3}{2} \Big)^{2} \\\\\\2x < 2 \ \ \ \Rightarrow \ \ \ x < 1[/tex]
[tex]\displaystyle\bf\\x\in(-\infty \ ; \ 1)\\\\\\Otvet \ : \ 0[/tex]
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[tex]\displaystyle\bf\\2^{2x+1}-3^{2x+1} > 3^{2x} -7\cdot 2^{2x} \\\\\\2^{2x+1}+7\cdot 2^{2x} > 3^{2x+1} + 3^{2x} \\\\\\2^{2x} \cdot(2+7) > 3^{2x} \cdot(3+1)\\\\\\2^{2x} \cdot 9 > 3^{2x} \cdot 4 \ |: \ 2^{2x} > 0 \\\\\\\frac{2^{2x} \cdot 9}{2^{2x} } > \frac{3^{2x} \cdot 4}{2^{2x} } \\\\\\\Big(\frac{3}{2} \Big)^{2x} \cdot 4 < 9\\\\\\\Big(\frac{3}{2} \Big)^{2x} < \frac{9}{4} \\\\\\\Big(\frac{3}{2} \Big)^{2x} < \Big(\frac{3}{2} \Big)^{2} \\\\\\2x < 2 \ \ \ \Rightarrow \ \ \ x < 1[/tex]
[tex]\displaystyle\bf\\x\in(-\infty \ ; \ 1)\\\\\\Otvet \ : \ 0[/tex]