Question

помогите решить срочно y+3/y-3-y-3/y+3 12y/y 2+6y+9

Answer 1

$\frac{y+3}{y-3} - \frac{y-3}{y+3} = \frac{(y+3)(y+3)-(y-3)(y-3)}{(y-3)(y+3)} = \frac{ (y+3)^{2}-(y-3)^{2} }{ y^{2} -9} = \\ \\ \frac{(y+3+y-3)(y+3-y+3)}{y^{2} -9} = \frac{2y*6}{y^{2} -9} = \frac{12y}{y^{2} -9} \\ \\ \\ \frac{12y}{y^{2} +6y+9} = \frac{12y}{(y+3)^{2} }$