Đáp án: $\dfrac{1}{x}$
Giải thích các bước giải:
Đặt $A=\dfrac{a}{x(x+a)}+\dfrac{a}{(x+a)(x+2a)}+\dfrac{a}{(x+2a)(x+3a)}+...+\dfrac{a}{(x+9a)(x+10a)}+\dfrac{1}{x+10a}$
Ta có:
$\dfrac{a}{x(x+a)}=\dfrac{1}{x}-\dfrac{1}{x+a}$
$\dfrac{a}{(x+a)(x+2a)}=\dfrac{1}{x+a}-\dfrac{1}{x+2a}$
$\dfrac{a}{(x+2a)(x+3a)}=\dfrac{1}{x+2a}-\dfrac{1}{x+3a}$
...
$\dfrac{a}{(x+9a)(x+10a)}=\dfrac{1}{x+9a}-\dfrac{1}{x+10a}$
Như vậy ta có:
$A=\dfrac{1}{x}-\dfrac{1}{x+a}+\dfrac{1}{x+a}-\dfrac{1}{x+2a}+\dfrac{1}{x+2a}-\dfrac{1}{x+3a}+...+\dfrac{1}{x+9a}-\dfrac{1}{x+10a}+\dfrac{1}{x+10a}$
$=\dfrac{1}{x}$