A = $\frac{x^{2} }{5x + 25}$ + $\frac{2(x-5)}{x}$ + $\frac{50 + 5x}{x(x+5)}$ (x $\neq$ 0 ;-5)
=$\frac{x^{2} }{5(x+5)}$ + $\frac{2(x-5)}{x}$ + $\frac{5(10+x)}{x(x+5)}$
=$\frac{x³ }{5x(x+5)}$ + $\frac{10(x-5)(x+5)}{5x(x+5)}$ + $\frac{25(10+x)}{5x(x+5)}$
=$\frac{x³+ 10(x²-25)+250+25x }{5x(x+5)}$
=$\frac{x³+ 10x²-250+250+25x }{5x(x+5)}$
=$\frac{x³+ 10x²+25x }{5x(x+5)}$
=$\frac{x(x²+10x+25) }{5x(x+5)}$
=$\frac{x(x+5)² }{5x(x+5)}$
=$\frac{x+5 }{5}$
Thay x = -2 vào A có A = $\frac{-2+5 }{5}$=$\frac{3 }{5}$
