Ta có :
$\dfrac{1}{a}.\left ( \dfrac{1}{x}-\dfrac{1}{x+a} \right )$
$=\dfrac{1}{a}.\left [ \dfrac{x+a}{x(x+a)}-\dfrac{x}{x(x+a)} \right ]$
$=\dfrac{1}{a}.\dfrac{x+a-x}{x(x+a)}$
$=\dfrac{1}{a}.\dfrac{a}{x(x+a)}$
$=\dfrac{1}{x(x+a)}$ ( ĐPCM )
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