$\frac{1}{x+1}$ - $\frac{5}{x-2}$ = $\frac{15}{(x+1)(2-x)}$  (ĐKXĐ: x $\neq$ -1; x $\neq$ 2)

$\frac{1}{x+1}$ + $\frac{5}{2-x}$ = $\frac{15}{(x+1)(2-x)}$ 

$\frac{2-x}{(x+1)(2-x)}$ + $\frac{5(x+1)}{(x+1)(2-x)}$ = $\frac{15}{(x+1)(2-x)}$ 

$\frac{2-x+5x+5}{(x+1)(2-x)}$ = $\frac{15}{(x+1)(2-x)}$ 

2 - x + 5x + 5 = 15

7 + 4x = 15

       4x = 15 - 7

       4x = 8

         x = 8 : 4

         x = 2 (KTM)

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