$\begin{align}  & {{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{2}}-{{\left( \,\sqrt{x}-\frac{1}{\sqrt{x}} \right)}^{2}} \\ & {{a}^{2}}-{{b}^{2}}=(a+b)(a-b) \\ & {{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{2}}-{{\left( \,\sqrt{x}-\frac{1}{\sqrt{x}} \right)}^{2}}=\left( \sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{x}-\frac{1}{\sqrt{x}} \right)\left( \sqrt{x}+\frac{1}{\sqrt{x}}-\left( \,\sqrt{x}-\frac{1}{\sqrt{x}} \right) \right) \\ & {{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{2}}-{{\left( \,\sqrt{x}-\frac{1}{\sqrt{x}} \right)}^{2}}=\left( \sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{x}-\frac{1}{\sqrt{x}} \right)\left( \sqrt{x}+\frac{1}{\sqrt{x}}-\sqrt{x}+\frac{1}{\sqrt{x}} \right) \\ & {{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{2}}-{{\left( \,\sqrt{x}-\frac{1}{\sqrt{x}} \right)}^{2}}=2\sqrt{x}\times \frac{2}{\sqrt{x}}=4 \\\end{align}$