\(PT↔\dfrac{1}{2x+1}+\dfrac{1}{2\sqrt{\left(2-x\right)\left(x+3\right)}}=\dfrac{7}{12}\)

ĐK: \(\begin{cases} -3<x<2\\ x\neq -\dfrac{1}{2}\end{cases}\)

Đặt \(\begin{cases} 2x+1=a\\ 2\sqrt{\left(2-x\right)\left(x+3\right)}=b\ge 0\end{cases}\)

\(\to a^2+b^2=4x^2+4x+1+24-4x-4x^2=25\)

Ta có hệ:

\(\begin{array}\\
\begin{cases}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{7}{12}\\ a^2+b^2=25\end{cases}\\
↔\begin{cases}12(a+b)=7ab\\ (a+b)^2-2ab=25\end{cases}\\
↔\begin{cases}S=\dfrac{7}{12}P\\ S^2-2P=25\end{cases}với\,\begin{cases}S=a+b\\ P=ab\end{cases}\left(S^2-4P\ge 0\right)\\
↔\begin{cases}S=\dfrac{7}{12}P(*)\\\dfrac{49}{144}P^2-2P=25(**)\end{cases}\\
(**)↔49P^2-288P-3600=0\\
↔(P-12)(49P+300)=0\\
\to \left[ \begin{array}{l}P=12\to S=7(n)\\P=-\dfrac{300}{49}\to S=-\dfrac{25}{7}(n)\end{array} \right.\\
TH_1:\begin{cases} P=ab=12\\ S=a+b=7 \end{cases}\\
\to a,b\,la\,nghiem\,cua\,PT\,m^2-7m+12=0\\
↔m=3\,or\,m=4\\
\to (a;b)=\left\{(4;3);(3;4)\right\}\\
Voi\,(a;b)=(4;3)\\
\to \begin{cases} 2x+1=4\\ 2\sqrt{\left(2-x\right)\left(x+3\right)}=3\end{cases}↔x=\dfrac{3}{2}(n)\\
Voi\,(a;b)=(3;4)\\
\to \begin{cases} 2x+1=3\\ 2\sqrt{\left(2-x\right)\left(x+3\right)}=4\end{cases}↔x=1(n)\\
TH_2:\begin{cases} P=ab=-\dfrac{300}{49}\\S=a+b=-\dfrac{25}{7} \end{cases}\\
\to a,b\,la\,nghiem\,cua\,PT\,n^2+\dfrac{25}{7}n-\dfrac{300}{49}=0\\
↔n=\dfrac{-25+5\sqrt{73}}{14}\,or\,n=\dfrac{-25-5\sqrt{73}}{14}\\
\to (a;b)=\left(\dfrac{-25-5\sqrt{73}}{14};\dfrac{-25+5\sqrt{73}}{14}\right)(b\ge 0)\\
\to \begin{cases}2x+1=\dfrac{-25-5\sqrt{73}}{14}\\ 2\sqrt{(2-x)(x+3)}=\dfrac{-25+5\sqrt{73}}{14}\end{cases}↔x=\dfrac{-39-5\sqrt{73}}{28}(n)
\end{array}\)

Kết luận: \(S=\left\{1;\dfrac{3}{2};\dfrac{-39-5\sqrt{73}}{28}\right\}\)