Answer
$\frac{12}{5}$
Work Step by Step
$16{{\left( \frac{x-1}{x-8} \right)}^{2}}+8\left( \frac{x-1}{x-8} \right)+1=0$ …… (1)
Let $u=\frac{x-1}{x-8}$ and ${{u}^{2}}={{\left( \frac{x-1}{x-8} \right)}^{2}}$
Substitute the values of $u$ and ${{u}^{2}}$ in equation (1),
$16{{u}^{2}}+8u+1=0$
Now, factor the equation,
$\left( 4u+1 \right)\left( 4u+1 \right)=0$
$4u+1=0$
$4u=-1$
Divide both sides by $4$,
$u=-\frac{1}{4}$
Now, replace $u$ with $\frac{x-1}{x-8}$
So, $\frac{x-1}{x-8}=-\frac{1}{4}$
Multiply both sides by $4\left( x-8 \right)$,
$\begin{align}
& 4\left( x-1 \right)=-\left( x-8 \right) \\
& 4x-4=-x+8 \\
& 4x+x=8+4 \\
& 5x=12
\end{align}$
Divide both the sides by $5$,
$x=\frac{12}{5}$
