Question

c+d/3-2c:c²+2cd+d²/2c²-3c​

Answer 1

   $\frac{c+d}{3-2c} :\frac{c^2+2cd+d^2}{2c^2-3c} =$

$=\frac{c+d}{3-2c} :\frac{(c+d)^2}{c(2c-3)} =$

$=\frac{c+d}{3-2c} *\frac{c(2c-3)}{(c+d)^2} =$

$=-\frac{c+d}{2c-3} *\frac{c(2c-3)}{(c+d)(c+d)} =$

$=-\frac{c}{c+d}$