giup em câu này với ạa,e cần gấp (e cảm ơn ạa)

Ta có: $\frac{x+3}{197}$ + $\frac{x+2}{198}$ + $\frac{x+1}{199}$ = $\frac{x+99}{101}$ + $\frac{x+98}{102}$ + $\frac{x+97}{103}$

$\frac{x+3}{197}$ + 1 + $\frac{x+2}{198}$ + 1 + $\frac{x+1}{199}$ + 1 = $\frac{x+99}{101}$ + 1 + $\frac{x+98}{102}$ + 1 + $\frac{x+97}{103}$ + 1

$\frac{x+3+197}{197}$ + $\frac{x+2+198}{198}$ + $\frac{x+1+199}{199}$ = $\frac{x+99+101}{101}$ + $\frac{x+98+102}{102}$ + $\frac{x+97+103}{103}$

$\frac{x+200}{197}$ + $\frac{x+200}{198}$ + $\frac{x+200}{199}$ - $\frac{x+200}{101}$ - $\frac{x+200}{102}$ - $\frac{x+200}{103}$ = 0

$(x+200)(\frac{1}{197}+\frac{1}{198}+\frac{1}{199}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103})$ = 0

Vì $\frac{1}{197}+\frac{1}{198}+\frac{1}{199}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103} $$\neq 0$ nên x + 200 = 0

⇒ x = -200

Vậy phương trình có nghiệm x = -200