Question

Срочно
$\frac{ {4a}^{2} - 1 }{ {a}^{2} - 9 } \div \frac{6a + 3}{a + 3}$
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Answer 1

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