ابتدا توجه کنید که $\sqrt{3}\approx 1/7$ پس: $\begin{align}  & f(x)={{x}^{2}}-2\left[ x \right] \\  & \Rightarrow f(\sqrt{3})={{(\sqrt{3})}^{2}}-2\left[ \sqrt{3} \right]=3-2\times 1=1 \\  & \Rightarrow -\frac{1}{2}f(\sqrt{3})=-\frac{1}{2}\times 1=-0/5 \\  & \Rightarrow f(-\frac{1}{2}f(\sqrt{3}))={{(-0/5)}^{2}}-2\left[ -0/5 \right] \\  & \Rightarrow f(-\frac{1}{2}f(\sqrt{3}))=0/25-2(-1)=2/25 \\ \end{align}$