$\text{Đáp án:ĐKXĐ $x\neq10;12$}$
$\dfrac{5}{x-10}-\dfrac{6}{x-12}=1$
$⇔\dfrac{5(x-12)-6(x-10)}{(x-10)(x-12)}=1$
$⇔5x-60-6x+60=(x-10)(x-12)$
$⇔-x=x^2-12x-10x+120$
$⇔x^2-21x+120=0$
$⇔x^2-\dfrac{21}{2}.2.x+(\dfrac{21}{2})^2-(\dfrac{21}{2})^2+120=0$
$⇔(x-\dfrac{21}{2})^2+\dfrac{39}{4}=0(vô lí)$
$\text{Vì $(x-\dfrac{21}{2})^2+\dfrac{39}{4}≥\dfrac{39}{4}>0$ }$
$\text{⇒Phương trình vô nghiệm}$
