4. The equation
$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$

has the solution(s) ( ).
(A) no solution
(B) exactly one solution
(C) exactly two solutions
(D) infinitely many solutions
4. D.

Transform the equation into
$$
\sqrt{(\sqrt{x-1}-2)^{2}}+\sqrt{(\sqrt{x-1}-3)^{2}}=1 \text {. }
$$

Consider three cases.
(1) $\sqrt{x-1} \geqslant 3$.

Then equation (1) becomes
$$
(\sqrt{x-1}-2)+(\sqrt{x-1}-3)=1 \text {, }
$$

which implies $\sqrt{x-1}=3 \Rightarrow x=10$.
(2) $\sqrt{x-1} \leqslant 2$.
Then equation (1) becomes
$$
(2-\sqrt{x-1})+(3-\sqrt{x-1})=1 \text {, }
$$

which implies $\sqrt{x-1}=2 \Rightarrow x=5$.
(3) $2<\sqrt{x-1}<3$, i.e., $5<x<10$.
Then equation (1) becomes
$$
(\sqrt{x-1}-2)+(3-\sqrt{x-1})=1 \text {, }
$$

which is an identity, meaning any $x$ satisfying $5<x<10$ is a solution to the equation.

In summary, the solutions to the equation are all real numbers satisfying $5 \leqslant x \leqslant 10$, i.e., there are infinitely many solutions.