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Решите задачу:

$1)\frac{x^{2}-y^{2} }{3x-3y } =\frac{(x-y)*(x+y)}{3(x-y)}=\frac{x+y}{3};\\ 2) (\frac{5y}{y}-\frac{x}{5y}):\frac{x}{y} =(5-\frac{x}{5y})*\frac{y}{x}=\frac{25y-x}{5y}*\frac{y}{x}=\frac{25y-x}{5}*\frac{1}{x}=\frac{25y-x}{5x}\\$

$3)(4x-\frac{12x}{x-2}) : (x-\frac{8x-25}{x-2})=\frac{4x*(x-2)-12x}{x-2}:\frac{x*(x-2)-(8x-25)}{x-2}=\frac{4x^{2}-8x-12x }{x-2} :\frac{x^{2}-2x-8x+25 }{x-2}=\frac{4x^2-20x}{x-2}*\frac{x^2-10x+25}{x-2}=\frac{4x*(x-5)}{x-2} *\frac{x-2}{x^2-10x+25}=4x*(x-5)*\frac{1}{(x-5)^{2} } =4x*\frac{1}{x-5}=\frac{4x}{x-5}$