Đáp án: $x=10$

Giải thích các bước giải:

Ta có:

$\dfrac{x-1}9+\dfrac{x-2}8+\dfrac{x-3}7=\dfrac{x-4}6+\dfrac{x-5}5+\dfrac{x-6}4$

$\to \dfrac{x-1}9-1+\dfrac{x-2}8-1+\dfrac{x-3}7-1=\dfrac{x-4}6-1+\dfrac{x-5}5-1+\dfrac{x-6}4-1$

$\to \dfrac{x-10}9+\dfrac{x-10}8+\dfrac{x-10}7=\dfrac{x-10}6+\dfrac{x-10}5+\dfrac{x-10}4$

$\to \dfrac{x-10}9+\dfrac{x-10}8+\dfrac{x-10}7-\dfrac{x-10}6-\dfrac{x-10}5-\dfrac{x-10}4=0$

$\to (x-10)\cdot (\dfrac19+\dfrac18+\dfrac17-\dfrac16-\dfrac15-\dfrac14)=0$

$\to x-10=0$

$\to x=10$