Explanation

Given, $U = \frac{A\sqrt{x}}{x + B}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$ Dimensions of $U =$ dimensions of potential energy $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left\lbrack ML^{2}T^{- 2} \right\rbrack$ From Eq. (i), Dimensions of $B = \$dimensions of $x = \left\lbrack M^{0}LT^{0} \right\rbrack$ $\therefore$ Dimensions of $A$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{dimensions\ of\ U\ \times dimensions\ of\ (x + B)}{dimension\ of\ \sqrt{x}}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\left\lbrack ML^{2}T^{- 2} \right\rbrack\left\lbrack M^{0}LT^{0} \right\rbrack}{\left\lbrack M^{0}L^{1/2}T^{0} \right\rbrack}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left\lbrack ML^{5/2}T^{- 2} \right\rbrack$ Hence, dimensions of $AB$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left\lbrack ML^{5/2}T^{- 2} \right\rbrack\left\lbrack M^{0}LT^{0} \right\rbrack$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left\lbrack ML^{7/2}T^{- 2} \right\rbrack$