a: ĐKXĐ: x<>0; x<>-5
b: \(A=\dfrac{x^3+2x^2+2\left(x-5\right)\left(x+5\right)+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}=\dfrac{\left(x+5\right)\left(x-1\right)\cdot x}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
b: để A=1 thì x-1=2
=>x=3(nhận)
Để A=-3 thì x-1=-6
=>x=-5(loại)