Question

помогите пожалуйста срочно пожалуйста​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{m}{x-y} +\frac{2m}{3(x-y)} =\frac{3m+2m}{3(x-y)} =\frac{5m}{3(x-y)} \\\\2)\\\\\frac{7a}{2(a-1)} -\frac{5a^{2} }{4(a-1)} =\frac{7a\cdot 2-5a^{2} }{4(a-1)} =\frac{14a-5a^{2} }{4(a-1)} \\\\3)\\\\\frac{2a^{2} }{3(a+1)} +\frac{5a^{2} }{4(a+1)} =\frac{2a^{2}\cdot 4+5a^{2} \cdot 3 }{3\cdot 4\cdot(a+1)} =\frac{8a^{2} +15a^{2} }{12(a+1)} =\frac{23a^{2} }{12(a+1)} \\\\4)$

$\displaystyle\bf\\\frac{4y}{5(y-3)} -\frac{5y}{2(y-3)} =\frac{4y\cdot2-5y\cdot 5}{5\cdot 2\cdot(y-3)} =\frac{8y-25y}{10(y-3)} =\frac{17y}{10(3-y)}$