The potential energy of a particle changes with distance x from a...

By the principle of dimensional homogeneity, quantities added or subtracted must have the same dimensions. In the denominator, $B$ is added to $x$ (distance).

$[B] = [x] = [L]$

The dimensions of potential energy $V$ are $[M^1 L^2 T^{-2}]$.

From the given equation $V = \dfrac{A\sqrt{x}}{x + B}$, we can write the dimensional formula as:

$[V] = \dfrac{[A][x]^{1/2}}{[x + B]}$

$[M^1 L^2 T^{-2}] = \dfrac{[A][L]^{1/2}}{[L]}$

$[M^1 L^2 T^{-2}] = [A][L]^{-1/2}$

$[A] = [M^1 L^2 T^{-2}][L]^{1/2} = [M^1 L^{5/2} T^{-2}]$

Now, the dimensions of $AB$ are:

$[AB] = [A][B] = [M^1 L^{5/2} T^{-2}][L] = [M^1 L^{7/2} T^{-2}]$

Answer: $[M^1 L^{7/2} T^{-2}]$