Question

ПОМОГИТЕ ПРОШУДАЮ 50 БАЛЛОВ ​

Answer 1

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

$4)\dfrac{a-5}{a(a+3)} -\dfrac{a+5}{a(a-3)}+\dfrac{16}{(a-3)^{2} } =\\\\=\dfrac{(a-5)\cdot(a-3)^{2}-(a+5)\cdot(a-3)(a+3)+16\cdot a(a+3) }{a(a+3)(a-3)^{2} } =\\\\=\dfrac{a^{3}-6a^{2}+9a-5a^{2}+30a-45-a^{3}+9a-5a^{2} +45+16a^{2}+48a}{a(a+3)(a-3)^{2} }=\\\\=\dfrac{96a}{a(a+3)(a-3)^{2} } =\dfrac{96}{(a+3)(a-3)^{2} }$

Answer 2

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