The set of all $\mathrm{a}\in \mathrm{ℝ}$ for which the equation $\mathrm{x}|\mathrm{x}-1|+|\mathrm{x}+2|+\mathrm{a}=0$ has exactly one real root, is

$x|x-1|+|x+2|+a=0$
$\text{ Case I: }x<-2$
$-x^2+x-x-2+a=0$
$a=x^2+2$
$y=x^2+2\text{ is decreasing }\mathrm{\forall }x\in (-\mathrm{\infty },-2)$
$\text{ Case II : }-2\le x<1$
$\begin{array}{l}-x^2+x+x+2+a=0\\ a=x^2-2x-2\end{array}$
$y=x^2-2x-2\text{ is decreasing }\mathrm{\forall }x\in [-2,1)$
$\text{ Case III : }x\ge 1$
$\begin{array}{l}{x}^2-x+x+2+a=0\\ a=-\left(x^2+2\right)\end{array}$
$y=-\left(x^2+2\right)\text{ is decreasing }\mathrm{\forall }x\in [1,\mathrm{\infty })$
$\therefore \text{ Exactly one real root }\mathrm{\forall }x\in R$