Question

9b²-1
-------------
9b²+6b+1​

Answer 1

Решение и ответ:

$\displaystyle \frac{{9b{}^2-1}}{{9{b^2}+6b+1}}=\frac{{{3^2}b{}^2-{1^2}}}{{{3^2}{b^2}+2\cdot 3\cdot1b+{1^2}}}=\frac{{{{\left({3b}\right)}^2}-{1^2}}}{{{{\left({3b}\right)}^2}+2\cdot 3b\cdot 1+{1^2}}}=\frac{{\left({3b-1}\right)\left({3b+1}\right)}}{{{{\left({3b+1}\right)}^2}}}=\frac{{3b-1}}{{3b+1}}$