Bài giải:

$\frac{5}{x}+\frac{4}{x+1}=\frac{3}{x+2}+\frac{2}{x+3}$

ĐKXĐ: x $\ne $0; x $\ne $-1; x $\ne $-2; x $\ne $-3

$\Leftrightarrow \frac{5}{x}+1+\frac{4}{x+1}+1=\frac{3}{x+2}+1+\frac{2}{x+3}+1$

$\Leftrightarrow \frac{5+x}{x}+\frac{4+x+1}{x+1}=\frac{3+x+2}{x+2}+\frac{2+x+3}{x+3}$

$\Leftrightarrow \frac{5+x}{x}+\frac{5+x}{x+1}=\frac{x+5}{x+2}+\frac{x+5}{x+3}$

$\Leftrightarrow (x+5)\left( \frac{1}{x}+\frac{1}{x+1}-\frac{1}{x+2}-\frac{1}{x+3} \right)=0$

Vì $\frac{1}{x}+\frac{1}{x+1}-\frac{1}{x+2}-\frac{1}{x+3}\ne 0$

Suy ra: x + 5 = 0

$\Leftrightarrow $ x = -5

Vậy x = -5