For $\mathrm{x}\in \mathrm{ℝ}$ the number of real roots of the equation  $3{\mathrm{x}}^2-4\left|{\mathrm{x}}^2-1\right|+\mathrm{x}-1=0$ is

$3x^2-4\left|x^2-1\right|+x-1=0$

$3x^2+x-1=4\left|x^2-2\right|$

Number of solutions is same as number of points of intersection of ${\text{y=3x}}^2+x-1 and y=4\left|x^2-2\right|$

the graphs intersect at 4 points. Therefore number of solutions=4

or

$3x^2-4\left|x^2-1\right|+x-1=0$

$\text{ Case 1: }{x}^2-1<0\Rightarrow x\in (-1,1)$

$\therefore 3x^2+4\left(x^2-1\right)+x-1=0$

$\Rightarrow 7x^2+x-5=0$

$\Rightarrow x=\frac{-1\pm \sqrt{141}}{14}$

$\Rightarrow x=\frac{-1-\sqrt{141}}{14},\frac{-1+\sqrt{141}}{14}$

$\text{ Both roots lie in }(-1,1)$

$\text{ Case }2:x^2-1\ge 0\Rightarrow x\in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$

$\therefore 3x^2-4\left(x^2-1\right)+x-1=0$

$\Rightarrow x^2-x-3=0$

$\Rightarrow x=\frac{1\pm \sqrt{13}}{2}$

$\Rightarrow x=\frac{1-\sqrt{13}}{2},\frac{1+\sqrt{13}}{2}$

$\text{ Both roots lie in }(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$

 Therefore number of solutions=4