Verified answer

[tex]\displaystyle\bf\\1)\\\\\frac{6a}{2a+5} -\frac{16a}{4a^{2} +20a+25} =\frac{6a}{2a+5}-\frac{16a}{(2a+5)^{2} } =\\\\\\=\frac{6a\cdot(2a+5)-16a}{(2a+5)^{2} }=\frac{12a^{2} +30a-16a}{(2a+5)^{2} } =\\\\\\=\frac{12a^{2}+14a }{(2a+5)^{2} }=\frac{2a\cdot(6a+7)}{(2a+5)^{2} } \\\\\\2)\\\\\frac{2a\cdot(6a+7)}{(2a+5)^{2} } :\frac{6a+7}{4a^{2} -25} =\frac{2a\cdot(6a+7)}{(2a+5)^{2} }\cdot\frac{4a^{2} -25}{6a+7} =[/tex]

[tex]\displaystyle\bf\\=\frac{2a\cdot(6a+7)}{(2a+5)^{2} } \cdot\frac{(2a+5)(2a-5)}{6a+7} =\frac{2a\cdot(2a-5)}{2a+5} =\frac{4a^{2} -10a}{2a+5} \\\\\\3)\\\\\frac{4a^{2} -10a}{2a+5} +\frac{10a-25}{2a+5} =\frac{4a^{2} -10a+10a-25}{2a+5} =\frac{4a^{2} -25}{2a+5} =\\\\\\=\frac{(2a-5)(2a+5)}{2a+5} =2a-5\\\\\\2a-5=2a-5[/tex]

Что и требовалось доказать