5. If $a_{0}+a_{1}(2 x-1)+a_{2}(2 x-1)^{2}+a_{3}(2 x-1)^{3}+a_{4}(2 x-1)^{4}+a_{5}(2 x-1)^{5}=x^{5}$, then $a_{2}=$
$$
x=\frac{1}{2}[(2 x-1)+1] \Rightarrow x^{5}=\frac{1}{32}[(2 x-1)+1]^{5} \text {, so } a_{2}=\frac{1}{32} C_{5}^{3}=\frac{5}{16} \text {. }
$$