Đáp án + Giải thích các bước giải:
$ĐKXĐ: x^{2}+6x\neq0⇔x(x+6)\neq0$ $⇔\left \{ {{x\neq0} \atop {x\neq-6}} \right.$
$\dfrac{14}{x}-\dfrac{2x}{x+6}+\dfrac{72}{x^{2}+6x}$
$=\dfrac{14(x+6)}{x(x+6)}-\dfrac{2x.x}{x(x+6)}+\dfrac{72}{x(x+6)}$
$=\dfrac{14x+84-2x^{2}+72}{x(x+6)}$
$=\dfrac{-2x^{2}+14x+156}{x(x+6)}$
$=\dfrac{-2(x^{2}-7x-78)}{x(x+6)}$
$=\dfrac{-2(x^{2}+6x-13x-78)}{x(x+6)}$
$=\dfrac{-2[x(x+6)-13(x+6)]}{x(x+6)}$
$=\dfrac{-2(x+6)(x-13)}{x(x+6)}$
$=\dfrac{-2(x-13)}{x}$
$=\dfrac{-2x+26}{x}$