Question

va rog ajutati-ma!!!!

Answer 1

j).
$\frac{y+2}{4y+6} : \frac{2y+4}{2y+3} = \frac{y+2}{2(2y+3)}* \frac{2y+3}{2y+4} = \frac{y+2}{2}* \frac{1}{2(y+2)} = \frac{1}{2}* \frac{1}{2} = \frac{1}{4}$
k).
$\frac{n+1}{n+2} * \frac{n+2}{n+3} : \frac{n+1}{n+3} = (n+1)* \frac{1}{n+3} * \frac{n+3}{n+1}= 1$
l).
[tex] \frac{2m+3}{6p+4}: \frac{4m+6}{9p+6} + \frac{3n+3}{8q+20}* \frac{2q+1}{3(n+1)}= \\ \\ = \frac{2m+3}{2(3p+2)} * \frac{9p+6}{4m+6} + \frac{3(n+1)}{8q+20} * \frac{2q+1}{3(n+1)} = \\ \\ = \frac{2m+3}{2(3p+2)} * \frac{3(3p+2)}{2(2m+3)} + \frac{1}{8q+20} *(2q+1) = \\ \\ = \frac{1}{2}* \frac{3}{2} + \frac{2q+1}{8q+20} = \frac{3}{4}+ \frac{2q+1}{4(2q+5)} = \\ \\ = \frac{3(2q+5)+2q+1}{4(2q+5)} = \frac{6q+15+2q+1}{4(2q+5)} = \\ = \frac{8q+16}{4(2q+5)}= \frac{4(2q+4)}{4(2q+5)}=\frac{2q +4}{2q+5} [/tex]