The number of real roots of the equation $\mathrm{x}\left|\mathrm{x}\right|-5\left|\mathrm{x}+2\right|+6=0$ is

$$\begin{gathered} x\left|x\right|-5\left|x+2\right|+6=0 \\ \begin{array}{l}case1: \\ If x<-2\\ \Rightarrow -x^2+5(x+2)+6=0\\ \Rightarrow x^2-5x-16=0\\ \Rightarrow x=\frac{5\pm \sqrt{25+64}}{2}\\ \therefore x=\frac{5-\sqrt{89}}{2} \left(which is less than -2\right)\end{array} \end{gathered}$$
$$\begin{gathered} case 2: \\ \begin{array}{l}If -2\le x<0\\ \Rightarrow -x^2-5(x+2)+6=0\\ \Rightarrow x^2+5x+4=0\\ \Rightarrow (x+1)(x+4)=0\\ x=-1\text{ ∈[-2,0)}\end{array} \end{gathered}$$

$$\begin{gathered} case 3: \\ \begin{array}{l}If x\ge 0\\ \Rightarrow x^2-5(x+2)+6=0\\ \Rightarrow x^2-5x-4=0\\ x=\frac{5\pm \sqrt{25+16}}{2}\\ =\frac{5\pm \sqrt{41}}{2}\\ x=\frac{5+\sqrt{41}}{2}\text{ >0}\end{array} \end{gathered}$$
Number of real roots=3