Tính: $\frac{a}{x(x+a)}$ + $\frac{a}{(x+a)(x+2a)}$ + $\frac{a}{(x+2a)(x+3a)}$ +…+ $\frac{a}{(x+9a)(x+10a)}$ + $\frac{1}{x+10a}$

0 bình luận về “Tính: $\frac{a}{x(x+a)}$ + $\frac{a}{(x+a)(x+2a)}$ + $\frac{a}{(x+2a)(x+3a)}$ +…+ $\frac{a}{(x+9a)(x+10a)}$ + $\frac{1}{x+10a}$”

1. 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:

   $A=\dfrac{1}{x}-\dfrac{1}{x+a}+\dfrac{1}{x+a}-\dfrac{1}{x+2a}+\dfrac{1}{x+2a}-\dfrac{1}{x+3a}+…+\dfrac{1}{x+9a}-\dfrac{1}{x+10a}+\dfrac{1}{x+10a}$

   $=\dfrac{1}{x}$

   Bình luận