Your hint will not work here

first off $\sqrt{a-2\sqrt{b}}=|\sqrt{c}-\sqrt{d}|$ $a-2\sqrt{b}=|c+d-2\sqrt{dc}|$ $a=c+d,b=cd$ so, using this: $\sqrt{x-1+4-2\sqrt{4(x-1)}}+\sqrt{x-1+9-2\sqrt{9(x-1)}}=|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$ so,if both are positive or negative, $2\sqrt{x-1}-5=1,-1\Longrightarrow x=5,10$ the 2nd case is that one is negative and the other positive $\sqrt{x-1}-2-\sqrt{x-1}+3=1$ its true for all x, but the inequities give us $\begin{cases} \sqrt{x-1}-2>0\Longrightarrow x>5\\ \sqrt{x-1}-3<0\Longrightarrow x<10\end{cases}$ $x=6,7,8,9$ next case is for x<1, meaning the square root unreal.but it would have an complex number inside th original square root. hence neglected.

Take sqrt(x-1) as t

Then you get the simple equation |t-2| + |t-3| = 1

Solve it to get the integer values 5,6,7,8,9 and 10

Here we can substitute $x$ as $sec^2\theta$ since we have the identity $1 + tan^2\theta = sec^2\theta$ Thus, the original equation will be $\sqrt { \sec { { ^{ 2 }{ \theta } } } +3-4\sqrt { \sec { { ^{ 2 }{ \theta } } } -1 } } -\sqrt { \sec { ^{ 2 }{ \theta } } +8-6\sqrt { \sec ^{ 2 }{ \theta } -1 } } = 1$ After some rearangements and manipulations we'll get $\left| \tan { \theta } -2 \right| +\left| \tan { \theta } -3 \right| = 1$ Subsitute $x$ back to the equation as $\tan { \theta } =\sqrt { x-1 }$ and solve for $x$ we'll get $x\in \left\{ 5,6,7,8,9,10 \right\}$ .