Ans: D
$\begin{array}{rcl}
\dfrac{\alpha}{1-x} & = & \dfrac{\beta}{x} \\
\alpha x & = & \beta(1-x) \\
\alpha x & = & \beta – \beta x \\
\alpha x – \beta x & = & \beta \\
x(\alpha – \beta) & = & \beta \\
x & = & \dfrac{\beta}{\alpha – \beta}
\end{array}$
$\begin{array}{rcl}
\dfrac{\alpha}{1-x} & = & \dfrac{\beta}{x} \\
\alpha x & = & \beta(1-x) \\
\alpha x & = & \beta – \beta x \\
\alpha x – \beta x & = & \beta \\
x(\alpha – \beta) & = & \beta \\
x & = & \dfrac{\beta}{\alpha – \beta}
\end{array}$
