Xét $\sqrt{x-1}+\sqrt{x+1}=\sqrt{7}$
Đặt $\left\{ \begin{align}
& a=\sqrt{x-1} \\
& b=\sqrt{x+1} \\
\end{align} \right.$ $\left( a\ge 0,\text{ }b\ge 0 \right)$
Khi đó ta có:
$\left\{ \begin{align}
& {{a}^{2}}-{{b}^{2}}=-2 \\
& a+b=\sqrt{7} \\
\end{align} \right.\Leftrightarrow \left\{ \begin{align}
& \left( a+b \right)\left( a-b \right)=-2 \\
& a+b=\sqrt{7} \\
\end{align} \right.\Leftrightarrow \left\{ \begin{align}
& a-b=\frac{-2}{\sqrt{7}} \\
& a+b=\sqrt{7} \\
\end{align} \right.\Leftrightarrow \left\{ \begin{align}
& a=\frac{5\sqrt{7}}{14} \\
& b=\frac{9\sqrt{7}}{14} \\
\end{align} \right.$Suy ra: $\sqrt{x-1}=\frac{5\sqrt{7}}{14}\Leftrightarrow x-1=\frac{25}{28}\Leftrightarrow x=\frac{53}{28}$ (Thỏa mãn $x\ge 1$)
Xét biểu thức: $A=\sqrt{2x+2\sqrt{{{x}^{2}}-1}}$.
Với $x=\frac{53}{28}$ ta có $A=\sqrt{2\cdot \frac{53}{28}+2\sqrt{{{\left( \frac{53}{28} \right)}^{2}}-1}}=\sqrt{\frac{53}{14}+2\sqrt{\frac{2809}{784}-1}}=\sqrt{\frac{53}{14}+2\sqrt{\frac{2025}{784}}}=\sqrt{\frac{53}{14}+2\cdot \frac{45}{28}}=\sqrt{7}$.

Xét $\sqrt{x-1}+\sqrt{x+1}=\sqrt{7}$
Đặt $\left\{ \begin{align}
& a=\sqrt{x-1} \\
& b=\sqrt{x+1} \\
\end{align} \right.$ $\left( a\ge 0,\text{ }b\ge 0 \right)$
Khi đó ta có:
$\left\{ \begin{align}
& {{a}^{2}}-{{b}^{2}}=-2 \\
& a+b=\sqrt{7} \\
\end{align} \right.\Leftrightarrow \left\{ \begin{align}
& \left( a+b \right)\left( a-b \right)=-2 \\
& a+b=\sqrt{7} \\
\end{align} \right.\Leftrightarrow \left\{ \begin{align}
& a-b=\frac{-2}{\sqrt{7}} \\
& a+b=\sqrt{7} \\
\end{align} \right.\Leftrightarrow \left\{ \begin{align}
& a=\frac{5\sqrt{7}}{14} \\
& b=\frac{9\sqrt{7}}{14} \\
\end{align} \right.$
Suy ra: $\sqrt{x-1}=\frac{5\sqrt{7}}{14}\Leftrightarrow x-1=\frac{25}{28}\Leftrightarrow x=\frac{53}{28}$ (Thỏa mãn $x\ge 1$)
Xét biểu thức: $A=\sqrt{2x+2\sqrt{{{x}^{2}}-1}}$.
Với $x=\frac{53}{28}$ ta có $A=\sqrt{2\cdot \frac{53}{28}+2\sqrt{{{\left( \frac{53}{28} \right)}^{2}}-1}}=\sqrt{\frac{53}{14}+2\sqrt{\frac{2809}{784}-1}}=\sqrt{\frac{53}{14}+2\sqrt{\frac{2025}{784}}}=\sqrt{\frac{53}{14}+2\cdot \frac{45}{28}}=\sqrt{7}$.

Xét $\sqrt{x-1}+\sqrt{x+1}=\sqrt{7}$
Đặt $\left\{ \begin{align}& a=\sqrt{x-1} \\ & b=\sqrt{x+1} \\\end{align} \right.$ $\left( a\ge 0,\text{ }b\ge 0 \right)$
Khi đó ta có:
$\left\{ \begin{align}& {{a}^{2}}-{{b}^{2}}=-2 \\ & a+b=\sqrt{7} \\\end{align} \right.\Leftrightarrow \left\{ \begin{align} & \left( a+b \right)\left( a-b \right)=-2 \\ & a+b=\sqrt{7} \\\end{align} \right.\Leftrightarrow \left\{ \begin{align}& a-b=\frac{-2}{\sqrt{7}} \\ & a+b=\sqrt{7} \\\end{align} \right.\Leftrightarrow \left\{ \begin{align}& a=\frac{5\sqrt{7}}{14} \\& b=\frac{9\sqrt{7}}{14} \\\end{align} \right.$
Suy ra: $\sqrt{x-1}=\frac{5\sqrt{7}}{14}\Leftrightarrow x-1=\frac{25}{28}\Leftrightarrow x=\frac{53}{28}$ (Thỏa mãn $x\ge 1$)
Xét biểu thức: $A=\sqrt{2x+2\sqrt{{{x}^{2}}-1}}$.
Với $x=\frac{53}{28}$ ta có $A=\sqrt{2\cdot \frac{53}{28}+2\sqrt{{{\left( \frac{53}{28} \right)}^{2}}-1}}=\sqrt{\frac{53}{14}+2\sqrt{\frac{2809}{784}-1}}=\sqrt{\frac{53}{14}+2\sqrt{\frac{2025}{784}}}=\sqrt{\frac{53}{14}+2\cdot \frac{45}{28}}=\sqrt{7}$.