$x^{2}$ + $\frac{9x^2}{( x + 3)^2}$ = 7 (1)

ĐK: x$\neq$ -3

 (1) <=> $\frac{x^2.(x+3)^2 +9x^2 - 7.(x+3)^2}{(x+3)^2}$ = 0

 <=> $x^2.(x+3)^2 +9x^2 - 7.(x+3)^2$ = 0

<=>$x^2.(x^2 +6x+9) +9x^2 - 7.(X^2 + 6x+9)$ =0

<=>$X^4 +   6x^3 + 9x^2 + 9x^2 -7x^2-42x - 63$=0

<=> $X^4 +   6x^3 + 11x^2 -42x-63 =0 $

<=> $(x^2 -x-3).(x^2 + 7x + 21) = 0$

<=> \(\left[ \begin{array}{l}x^2 -x-3 =0\\x^2 + 7x + 21 = 0 ( vô . nghiệm)\end{array} \right.\) 

 <=> x = $\frac{1 + \sqrt[2]{13} }{2}$  hoặc x = $\frac{1 - \sqrt[2]{13} }{2}$  

$\begin{array}{l} {x^2} + \dfrac{{9{x^2}}}{{{{\left( {x + 3} \right)}^2}}} = 7\\  \Leftrightarrow {\left( {x - \dfrac{{3x}}{{x + 3}}} \right)^2} + \dfrac{{2.3{x^2}}}{{x + 3}} = 7\\  \Leftrightarrow {\left( {\dfrac{{{x^2}}}{{x + 3}}} \right)^2} + \dfrac{{6{x^2}}}{{x + 3}} = 7\\  \Leftrightarrow {t^2} + 6t - 7 = 0\left( {t = \dfrac{{{x^2}}}{{x + 3}}} \right)\\  \Leftrightarrow \left[ \begin{array}{l} t =  - 7\\ t = 1 \end{array} \right. \Rightarrow \left[ \begin{array}{l} \dfrac{{{x^2}}}{{x + 3}} =  - 7\\ \dfrac{{{x^2}}}{{x + 3}} = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {x^2} + 7x + 21 = 0(PTVN)\\ {x^2} - x - 3 = 0 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{1 + \sqrt {13} }}{2}\\ x = \dfrac{{1 - \sqrt {13} }}{2} \end{array} \right. \Rightarrow S = \left\{ {\dfrac{{1 + \sqrt {13} }}{2};\dfrac{{1 - \sqrt {13} }}{2}} \right\} \end{array}$