根据题意，有\[f'(x)=\dfrac{2x-1}{\sqrt{2x^2-2x+1}}-\dfrac{2x+1}{\sqrt{2x^2+2x+5}}.\]记\[\begin{split} \varphi(x)&=\dfrac{1}{\sqrt{\dfrac{2x^2-2x+1}{4x^2-4x+1}}}-\dfrac{1}{\sqrt{\dfrac{2x^2+2x+5}{4x^2+4x+1}}}\\
&=\dfrac{1}{\sqrt{\dfrac 12+\dfrac 12\cdot \dfrac{1}{(2x-1)^2}}}-\dfrac{1}{\sqrt{\dfrac 12+\dfrac 12\cdot \dfrac{9}{(2x+1)^2}}}.\end{split}\]情形一当 $x>\dfrac 12$ 时，$f'(x)=\varphi(x)$，而\[\varphi(x)\begin{cases} \geqslant 0,&x\geqslant 1,\\ <0 ,&\dfrac 12<x<1.\end{cases}\]情形二当 $-\dfrac 12<x<\dfrac 12$ 时，$f'(x)<0$．
情形三当 $x<-\dfrac 12$ 时，$f'(x)=-\varphi(x)$，而此时 $\varphi(x)>0$．
又\[\lim_{x\to +\infty}f(x)=\dfrac{-4x-4}{\sqrt{2x^2-2x+1}+\sqrt{2x^2+2x+5}}=\dfrac{-4}{2\sqrt 2}=-\sqrt 2,\]且\[\lim_{x\to -\infty}f(x)=\dfrac{-4x-4}{\sqrt{2x^2-2x+1}+\sqrt{2x^2+2x+5}}=\dfrac{-4}{-2\sqrt 2}=\sqrt 2,\]因此\[\begin{array} {c|ccccc}\hline
x&-\infty&(-\infty,1)&1&(1,+\infty)&+\infty\\ \hline
f(x)&\sqrt 2&\searrow&-2&\nearrow &-\sqrt 2\\ \hline
\end{array}\]进而所求的值域为 $\left[-2,\sqrt 2\right)$．