Tìm số nguyên x thoả mãn

Giải thích các bước giải:

ĐKXĐ: $x\notin\{0, -2, -4, -6\}$

Ta có:

$\dfrac{2024}{x(x+2)}+\dfrac{2024}{(x+2)(x+4)}+\dfrac{2024}{(x+4)(x+6)}-\dfrac{1012}{x}=\dfrac12$

$\to 1012\cdot (\dfrac{2}{x(x+2)}+\dfrac{2}{(x+2)(x+4)}+\dfrac{2}{(x+4)(x+6)}-\dfrac{1}{x})=\dfrac12$

$\to 1012\cdot (\dfrac1x-\dfrac1{x+2}+\dfrac1{x+2}-\dfrac1{x+4}+\dfrac1{x+4}-\dfrac1{x+6}-\dfrac1x)=\dfrac12$

$\to 1012\cdot (-\dfrac1{x+6})=\dfrac12$

$\to x+6=-2024$

$\to x=-2030$