Verified answer

[tex]\displaystyle\bf\\1)\\\\\\\frac{2x-7}{2x+7} -\frac{2x+7}{2x-7} =\frac{(2x-7)\cdot(2x-7)-(2x+7)\cdot(2x+7)}{(2x+7)\cdot(2x-7)} =\\\\\\=\frac{4x^{2} -28x+49-4x^{2} -28x-49}{(2x+7)\cdot(2x-7)} =\frac{-56x}{(2x+7)\cdot(2x-7)} \\\\\\2)\\\\\\\frac{7x}{8x^{2} -98}: \frac{-56x}{(2x+7)\cdot(2x-7)} =\frac{7x}{2\cdot(4x^{2} -49)}: \frac{-56x}{(2x+7)\cdot(2x-7)} =\\\\\\=-\frac{7x}{2\cdot(2x+7)\cdot(2x-7)} \cdot \frac{(2x+7)\cdot(2x-7)}{56x} =-\frac{1}{2\cdot 8} =-\frac{1}{16}[/tex]

[tex]\displaystyle\bf\\-\frac{1}{16} =-\frac{1}{16}[/tex]

Что и требовалось доказать .