d: \(B=\dfrac{x^2-25+2x^2-12x-2x^2+2x+50}{2x\left(x-5\right)}\)
\(=\dfrac{\left(x-5\right)}{2x}\)
\(c,x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\\ \forall x=0\Leftrightarrow A=\dfrac{0-5}{0-4}=\dfrac{5}{4}\\ \forall x=3\Leftrightarrow A=\dfrac{3-5}{3-4}=2\\ d,B=\dfrac{x^2-25+2x^2-12x-2x^2+2x+50}{2x\left(x-5\right)}\\ B=\dfrac{x^2-10x+25}{2x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{2x\left(x-5\right)}=\dfrac{x-5}{2x}\\ c,P=A:B=\dfrac{x-5}{x-4}\cdot\dfrac{2x}{x-5}=\dfrac{2\left(x-4\right)+8}{x-4}=2+\dfrac{8}{x-4}\in Z\\ \Leftrightarrow x-4\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow x\in\left\{-4;2;3;6;8;12\right\}\)
