Matematika B - 2010. ljeto, 5.

$\frac{3a^2+6a+1}{9}\backslash frac\{3\; a^\{2\}+6\; a+1\}\{9\}$

$\frac{9a^2+6a+1}{9}\backslash frac\{9\; a^\{2\}+6\; a+1\}\{9\}$

$\frac{9a^2+3a+1}{3}\backslash frac\{9\; a^\{2\}+3\; a+1\}\{3\}$

$\frac{3a^2+3a+1}{3}\backslash frac\{3\; a^\{2\}+3\; a+1\}\{3\}$