Again, how many?

Noticed that $1-\sqrt{2(x^2-x+1)}<0; \forall x\in \mathbb{R}$ so in order for the inequality to happen $\begin{cases} x-\sqrt{x}<0\\ x-\sqrt{x}\leq 1-\sqrt{2(x^2-x+1)}\end{cases}$ $\begin{cases} x\in (0;1)\\ -x+\sqrt{x}+1\geq\sqrt{2(x^2-x+1)}\end{cases}$ Considering the second inequality we have $\sqrt{x}+(1-x)\geq\sqrt{2(x^2-x+1)}$ By Cauchy-Schwarz Inequality we have $LHS^2\leq 2[x+(x-1)^2]=2(x^2-x+1)=RHS^2$ $\Leftrightarrow LHS\leq RHS$ So $(1)$ happen only when $LHS=RHS$ $\Leftrightarrow\sqrt{x}=1-x$ $\Leftrightarrow x=\frac{3-\sqrt{5}}{2}$ Which satisfied $x\in (0;1)$ so $1$ is the correct choice

$\begin{aligned} \frac{x-\sqrt{x}}{\color{#D61F06}{1-\sqrt{2(x^2-x+1)}}} & \ge 1 \quad \quad \small \color{#D61F06}{\text{Since } 1-\sqrt{2(x^2-x+1)} < 0} \\ x-\sqrt{x} & \color{#D61F06}{\le} 1-\sqrt{2(x^2-x+1)} \\ x-1 & \le \sqrt{x} -\sqrt{2(x^2-x+1)} \quad \quad \small \color{#3D99F6}{\text{Squaring both sides}} \\ x^2 - 2x + 1 & \le x - 2\sqrt{2x(x^2-x+1)} + 2x^2-2x+2 \\ x^2 + x + 1 & \ge 2 \sqrt{2x(x^2-x+1)} \quad \quad \small \color{#3D99F6}{\text{Squaring both sides}} \\ x^4 + 2x^3 + 3x^2 + 2x +1 & \ge 8x^3 - 8x^2 + 8x \\ x^4 - 6x^3 + 11x^2 - 6x +1 &\ge 0 \quad \quad \small \color{#3D99F6}{\text{Dividing both sides by }x^2} \\ x^2 - 6x + 11 - \frac{6}{x} +\frac{1}{x^2} &\ge 0 \\ \left(x+\frac{1}{x}\right)^2 - 6\left(x+\frac{1}{x}\right) + 9 & \ge 0 \\ \left(x+\frac{1}{x} - 3 \right)^2 & \ge 0 \\ \Rightarrow x^2-3x+1 & \ge 0 \\ \left(x - \frac{3+\sqrt{5}}{2} \right)\left(x - \frac{3-\sqrt{5}}{2} \right) & \ge 0 \end{aligned}$

We note that:

$\begin{cases} x = \dfrac{3+\sqrt{5}}{2} & \Rightarrow \dfrac{x-\sqrt{x}}{1-\sqrt{2(x^2-x+1)}} \color{#D61F06} {< 1} \\ x = \dfrac{3-\sqrt{5}}{2} & \Rightarrow \dfrac{x-\sqrt{x}}{1-\sqrt{2(x^2-x+1)}} \color{#3D99F6} {= 1} \end{cases}$

Therefore, there is only $\boxed{1}$ value of $x$ satisfies the inequality.