Bài giải:

$\left( \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10} \right)(x-1)+\frac{1}{10}x=x-\frac{9}{10}$

$\Leftrightarrow \left( 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10} \right)(x-1)+\frac{1}{10}x=x-\frac{9}{10}$

$\Leftrightarrow \left( 1-\frac{1}{10} \right)(x-1)+\frac{1}{10}x=x-\frac{9}{10}$

$\Leftrightarrow \frac{9}{10}(x-1)+\frac{1}{10}x=x-\frac{9}{10}$

$\Leftrightarrow \frac{9}{10}x-\frac{9}{10}+\frac{1}{10}x=x-\frac{9}{10}$

$\Leftrightarrow \frac{9}{10}x+\frac{1}{10}x-x=-\frac{9}{10}+\frac{9}{10}$

$\Leftrightarrow $ 0x = 0

Phương trình có vô số nghiệm.

Vậy đáp án đúng là: D