Question

Прошу,кто поможет поставлю 5 звезд

Answer 1

$\displaystyle\bf\\1)\\\\\frac{7y+4}{8y}-\frac{2y+3}{6y} =\frac{(7y+4)\cdot 3-(2y+3)\cdot4}{24y} =\\\\\\=\frac{21y+12-8y-12}{24y} =\frac{13y}{24y} =\frac{13}{24} \\\\\\2)\\\\\frac{2-3y}{y^{2}-9 } -\frac{5-2y}{3-y} =\frac{2-3y}{(y-3)\cdot(y+3)}+\frac{5-2y}{y-3} =\\\\\\=\frac{2-3y+(5-2y)\cdot(y+3)}{(y-3)(y+3)} =\frac{2-3y+5y+15-2y^{2} -6y}{(y-3)(y+3)} =\\\\\\=\frac{-2y^{2}-4y+17 }{y^{2} -9}=-\frac{2y^{2} +4y-17}{y^{2} -9}$