Cho hai biểu thức:

\(A = \dfrac{x}{{x - 3}},\)\(B = \dfrac{{2x}}{{x + 5}} - \dfrac{{{x^2} - 15x}}{{{x^2} - 25}}\,\,\,\,\left( {x \ne 0;x \ne 3;x \ne  \pm 5} \right)\)

Rút gọn biểu thức \(Q = B:A\) ta được:

Điều kiện: \(x \ne 0,\,\,x \ne 3;\,\,x \ne  \pm 5.\)

\(B:A = \left( {\dfrac{{2x}}{{x + 5}} - \dfrac{{{x^2} - 15x}}{{{x^2} - 25}}} \right):\dfrac{x}{{x - 3}}\)\(= \left[ {\dfrac{{2x\left( {x - 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} - \dfrac{{{x^2} - 15x}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}} \right].\dfrac{{x - 3}}{x} \)\(= \dfrac{{2x\left( {x - 5} \right) - \left( {{x^2} - 15x} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}.\dfrac{{x - 3}}{x}\)\( = \dfrac{{2{x^2} - 10x - {x^2} + 15x}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}.\dfrac{{x - 3}}{x}\)\( = \dfrac{{{x^2} + 5x}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}.\dfrac{{x - 3}}{x}\)\( = \dfrac{{x\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}.\dfrac{{x - 3}}{x}\)\( = \dfrac{x}{{x - 5}}.\dfrac{{x - 3}}{x} = \dfrac{{x - 3}}{{x - 5}}\).

Vậy \(Q = B:A = \dfrac{{x - 3}}{{x - 5}}\).