If $\left(\mathrm{A}+\frac{\mathrm{B}}{{\mathrm{C}}^2}\right)(\mathrm{D}-\mathrm{x})=\mathrm{y}$, then 

$\left[\mathrm{D}\right]=\left[\mathrm{x}\right]\text{ and }\left[\mathrm{A}\right]=\left[\frac{\mathrm{B}}{{\mathrm{C}}^2}\right]$

The dimensions of A and B may be same if C is dimensionless. Hence option (2) may be correct.

$\begin{array}{l}\left(\mathrm{A}+\frac{\mathrm{B}}{{\mathrm{C}}^2}\right)(\mathrm{D}-\mathrm{x})=\mathrm{y}\\ \Rightarrow \mathrm{AD}-\mathrm{Ax}+\frac{\mathrm{BD}}{{\mathrm{C}}^2}-\frac{\mathrm{Bx}}{{\mathrm{C}}^2}=\mathrm{y}\\ \Rightarrow \left[\mathrm{y}\right]=\left[\mathrm{AD}\right]=\left[\mathrm{Ax}\right]=\left[\frac{\mathrm{BD}}{{\mathrm{C}}^2}\right]=\left[\frac{\mathrm{Bx}}{{\mathrm{C}}^2}\right]\end{array}$