Question

Допоможіть будь-ласка, 34 балла

$\frac{2x+y}{(x-y)^{2} } - \frac{x-4y}{(y-x)^{2} }$

$\frac{7a}{3(b+2a)} -\frac{4a}{9(b+2a)}$

$\frac{3n^{2}-8m^{2} }{n^{2}-2mn } - \frac{3mn-x^{2} }{mn-2m^{2} }$

Answer 1

$1)\frac{2x+y}{(x-y)^{2}}-\frac{x-4y}{(y-x)^{2}}=\frac{2x+y}{(x-y)^{2}}+\frac{x-4y}{(x-y)^{2}}=\frac{2x+y+x-4y}{(x-y)^{2}}=\frac{3x-3y}{(x-y)^{2}}=\frac{3(x-y)}{(x-y)(x+y)}=\frac{3}{x+y}$

$2)\frac{7a}{3(b+2a)}-\frac{4a}{9(b+2a)}=\frac{21a-4a}{9(b+2a)}=\frac{17a}{9(b+2a)}$

$3)\frac{3n^{2}-8m^{2}}{n^{2}-2mn}-\frac{3mn-n^{2}}{mn-2m^{2}}=\frac{3n^{2}-8m^{2}  }{n(n-2m)}-\frac{3mn-n^{2}}{m(n-2m)}=\frac{3n^{2}m-8m^{3}-3n^{2}m+n^{3}}{mn(n-2m)} =\frac{n^{3}-8m^{3}}{mn(n-2m)}=\frac{(n-2m)(n^{2}+mn+m^{2})}{mn(n-2m)}=\frac{n^{2}+mn+m^{2}}{mn}$