a: x^2-3x=0
=>x=0 hoặc x=3
Khi x=0 thì \(A=\dfrac{0-5}{0-4}=\dfrac{5}{4}\)
Khi x=3 thì \(A=\dfrac{3-5}{3-4}=\dfrac{-2}{-1}=2\)
b: \(B=\dfrac{x^2-25+2x^2-12x-2x^2+2x+50}{2x\left(x-5\right)}\)
\(=\dfrac{x^2-10x+25}{2x\left(x-5\right)}=\dfrac{x-5}{2x}\)