Ta có: \(B=\frac{x+5}{2x}-\frac{x+6}{5-x}-\frac{2x^2-2x-50}{2x^2-10x}\)
\(=\frac{x+5}{2x}+\frac{x+6}{x-5}-\frac{2x^2-2x-50}{2x\left(x-5\right)}\)
\(=\frac{\left(x+5\right)\left(x-5\right)}{2x\left(x-5\right)}+\frac{2x\left(x+6\right)}{2x\left(x-5\right)}-\frac{2x^2-2x-50}{2x\left(x-5\right)}\)
\(=\frac{x^2-25+2x^2+12x-2x^2+2x+50}{2x\left(x-5\right)}\)
\(=\frac{x^2+14x+25}{2x\left(x-5\right)}\)
Ta có: \(P=\frac{A}{B}\)
\(=\frac{x-5}{x-4}:\frac{x^2+14x+25}{2x\left(x-5\right)}\)
\(=\frac{\left(x-5\right)}{x-4}\cdot\frac{2x\left(x-5\right)}{x^2+14x+25}\)
\(=\frac{2x\left(x-5\right)^2}{\left(x-4\right)\left(x^2+14x+25\right)}\)
\(=\frac{2x\left(x^2-10x+25\right)}{x^3+14x^2+25x-4x^2-56x-100}\)
\(=\frac{2x^3-10x^2+50x}{x^3+10x^2-31x-100}\)
Tới đây bạn tự làm tiếp nhé
