Ответ:
ОДЗ: \(x\ne0\), \(x\ne5y\), \(x\ne-5y\), \(y\ne0\).
Разложим знаменатели и числитель последней дроби:
\[x^2-5xy=x(x-5y),\quad x^2+5xy=x(x+5y),\quad 25y^2-x^2=(5y-x)(5y+x).\]
\[\frac{x+5y}{x(x-5y)}-\frac{x-5y}{x(x+5y)}\]
\[=\frac{(x+5y)^2-(x-5y)^2}{x(x-5y)(x+5y)}=\frac{20xy}{x(x-5y)(x+5y)}=\frac{20y}{(x-5y)(x+5y)}.\]
Так как \((x-5y)(x+5y)=x^2-25y^2=-(25y^2-x^2)\),
\[\frac{20y}{(x-5y)(x+5y)}\cdot\frac{25y^2-x^2}{5y^2}=\frac{20y}{-(25y^2-x^2)}\cdot\frac{25y^2-x^2}{5y^2}=-\frac4y.\]
Ответ: \(-\frac4y\).
