Question

Simplificati fractiile, astfel încât sa devina ireductibile: 3a+6b supra 5a+10b...
9a+3 supra 12a

Answer 1

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

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

Answer 2

$\it \dfrac{3a + 6b}{5a+10b} = \dfrac{\ \ 3(a+2b)^{(a+2b}}{5(a+2b)}=\dfrac{3}{5} \\ \\ \\ \dfrac{9a+3}{12a} =\dfrac{\ \ 3(3a+1)^{(3}}{12a} = \dfrac{3a+1}{4a}$