Answer
$-\frac{3}{2}$
Work Step by Step
$9{{\left( \frac{x+2}{x+3} \right)}^{2}}-6\left( \frac{x+2}{x+3} \right)+1=0$ ……(1)
Let $u=\frac{x+2}{x+3}$ and ${{u}^{2}}={{\left( \frac{x+2}{x+3} \right)}^{2}}$
Substitute the values of $u$ and ${{u}^{2}}$ in equation (1),
$9{{\left( u \right)}^{2}}-6\left( u \right)+1=0$
Now, factor the above equation.
$\left( 3u-1 \right)\left( 3u-1 \right)=0$
$3u-1=0$
$3u=1$
$u=\frac{1}{3}$
Now, replace $u$ with $\frac{x+2}{x+3}$
So, $\frac{x+2}{x+3}=\frac{1}{3}$
Multiply both sides by $3\left( x+3 \right)$,
$\begin{align}
& 3\left( x+2 \right)=\left( x+3 \right) \\
& 3x+6=x+3 \\
& 3x-x=3-6 \\
& 2x=-3
\end{align}$
Divide both sides by $2$,
$x=-\frac{3}{2}$
