Question

дам 20 балов
$( \frac{2a}{a + 3} - \frac{4a}{ {a}^{2} + 6a + 9 }) \div \frac{a + 1}{ {a}^{2} + 9 }$
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Answer 1

$\left(\dfrac{2a}{a+3} -\dfrac{4a}{a^2+6a+9} \right):\dfrac{a+1}{a^2+9} =\left(\dfrac{2a(a+3)}{(a+3)^2} -\dfrac{4a}{(a+3)^2} \right)\cdot\dfrac{a^2+9}{a+1} =$

$=\dfrac{2a(a+3)-4a}{(a+3)^2} \cdot\dfrac{a^2+9}{a+1} =\dfrac{2a^2+6a-4a}{(a+3)^2} \cdot\dfrac{a^2+9}{a+1} =\dfrac{2a^2+2a}{(a+3)^2} \cdot\dfrac{a^2+9}{a+1} =$

$=\dfrac{2a(a+1)}{(a+3)^2} \cdot\dfrac{a^2+9}{a+1} =\dfrac{2a(a+1)(a^2+9)}{(a+3)^2(a+1)} =\boxed{\dfrac{2a(a^2+9)}{(a+3)^2}}$