Đáp án đúng: C
Giải chi tiết:Điều kiện: \(x \ne 0,\,\,x \ne 3;\,\,x \ne \pm 5.\)
\(\begin{array}{l}B:A = \left( {\frac{{2x}}{{x + 5}} - \frac{{{x^2} - 15x}}{{{x^2} - 25}}} \right):\frac{x}{{x - 3}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\frac{{2x\left( {x - 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} - \frac{{{x^2} - 15x}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}} \right].\frac{{x - 3}}{x} = \frac{{2x\left( {x - 5} \right) - \left( {{x^2} - 15x} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}.\frac{{x - 3}}{x}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2{x^2} - 10x - {x^2} + 15x}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}.\frac{{x - 3}}{x}\,\,\, = \frac{{{x^2} + 5x}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}.\frac{{x - 3}}{x}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{x\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}.\frac{{x - 3}}{x}\,\, = \frac{x}{{x - 5}}.\frac{{x - 3}}{x} = \frac{{x - 3}}{{x - 5}}.\end{array}\)
Vậy \(Q = B:A = \frac{{x - 3}}{{x - 5}}\) .
Chọn C.
