解 ：由绝对值的意义，得$x≥0$
当$0≤x≤1$时，$|1- \sqrt {(x-1)²}|=|1-(1-x)|=x$恒成立；
当$x＞1$时，$|1- \sqrt {(x-1)^2}|=|1-(x-1)|=|2-x|=x$，解得$x=1$，不符合题意，舍去
则$0≤x≤1$
∵$\sqrt {x^2+\frac {1}{4}-x}+ \sqrt {x²+\frac {1}{4}+x} =\sqrt {(x-\frac {1}{2})^2} + \sqrt {(x+\frac {1}{2})^2}$
∴原式$=|x-\frac {1}{2}|+x+\frac {1}{2}$
当$0≤x≤\frac {1}{2}$时，原式$=\frac {1}{2}-x+x+\frac {1}{2}=1$；
当$\frac {1}{2}＜x≤1$时，原式$=x-\frac {1}{2}+x+\frac {1}{2}=2x$
综上，$\sqrt {x²+\frac {1}{4}-x}+ \sqrt {x²+\frac {1}{4}+x}=\begin {cases}{1(0≤x≤\frac {1}{2})}\\{ 2x(\frac {1}{2}＜x≤1)}\end {cases}$
