Question

найди значение дроби 5/(9a-3)²+10​

Answer 1

$\frac{5}{(9a-3)^{2} }+10=\frac{5+10(9a-3)^{2} }{(9a-3)^{2} } =\frac{5+10(81a^{2}-54a+9) }{(9a-3)^{2} }=\frac{5+810a^{2}-540a+90 }{(9a-3)^{2} }=\frac{95+810a^{2}-540a }{(9a-3)^{2} } =\frac{810a^{2}-540a+95 }{(9a-3)^{2} }$