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Решите задачу:

$1)\frac{a^{2}-25}{a+3}*\frac{1}{a^{2} +5a}=\frac{(a-5)(a+5)}{a+3}*\frac{1}{a(a+5)}=\frac{a-5}{a(a+3)}\\\\2)\frac{a^{2}+10a+25 }{a^{3}-6a^{2}+9a}:\frac{a+5}{a-3}=\frac{(a+5)^{2} }{a(a^{2}-6a+9) }*\frac{a-3}{a+5}=\frac{(a+5)(a-3)}{a(a-3)^{2}}=\frac{a+5}{a(a-3)}\\\\3)(\frac{4}{a-3})^{2}=\frac{16}{(a-3)^{2} }$

$4)\frac{a-5}{a(a+3)}-\frac{a+5}{a(a-3)}=\frac{a^{2}-3a-5a+15-a^{2} -5a-3a-15}{a(a+3)(a-3)}=\frac{-16a}{a(a+3)(a-3)}=-\frac{16}{(a+3)(a-3)}\\\\5)-\frac{16}{(a+3)(a-3)}+\frac{16}{(a-3)^{2} }=\frac{-16a+48+16a+48}{(a+3)(a-3)^{2} }=\frac{96}{(a+3)(a-3)^{2} }$

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