Question

siplifiicati fractiile intrucat sa ajunga ireductibile: 3a+6b supra 5a+10b si 9a+4 supra 12a+4

Answer 1

$\frac{3a+6b}{5a+10b} = \frac{3(a+2b)}{5(a+2b)} = \frac{3}{5}$

$\frac{9a+4}{12a+4} = \frac{9a}{12a+4} + \frac{4}{12a+4} = \frac{9a}{4(3a+1)} + \frac{4}{4(3a+1)} = \frac{9a}{4(3a+1)} + \frac{1}{3a+1}$