3. Let the roots of the equation $(x-a)(x-b)-x=0$ be $c$ and $d$. Then the roots of the equation $(x-c)(x-d)+x=0$ are ( ).
(A) $a, b$
(B) $-a,-b$
(C) $c, d$
(D) $-c,-d$
3. A.

From the condition, we have
$$
\begin{array}{l}
(x-a)(x-\dot{b})-x=(x-c)(x-d) \\
\Rightarrow(x-c)(x-d)+x=(x-a)(x-b) .
\end{array}
$$

Therefore, the roots of $(x-c)(x-d)+x=0$ are $a$ and $b$.