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Решите задачу:

$\left(\frac{6a}{2a+5}-\frac{16a}{4a^2+20a+25}\right):\frac{6a+7}{4a^2-25}+\frac{10a-25}{2a+5}=$

$\left(\frac{6a}{2a+5}-\frac{16a}{(2a+5)^2}\right):\frac{6a+7}{(2a+5)(2a-5)}+\frac{10a-25}{2a+5}=$

$\left(\frac{6a(2a+5)}{(2a+5)^2}-\frac{16a}{(2a+5)^2}\right)\cdot\frac{(2a+5)(2a-5)}{6a+7}+\frac{10a-25}{2a+5}=$

$\frac{12a^2+30a-16a}{(2a+5)^2}\cdot\frac{(2a+5)(2a-5)}{6a+7}+\frac{10a-25}{2a+5}=$

$\frac{12a^2+14a}{2a+5}\cdot\frac{2a-5}{6a+7}+\frac{10a-25}{2a+5}=$

$\frac{2a(6a+7)}{2a+5}\cdot\frac{2a-5}{6a+7}+\frac{10a-25}{2a+5}=$

$\frac{2a(2a-5)}{2a+5}+\frac{10a-25}{2a+5}=$

$\frac{4a^2-10a+10a-25}{2a+5}=$

$\frac{4a^2-25}{2a+5}=$

$\frac{(2a-5)(2a+5)}{2a+5}=2a-5$