\(\sqrt{\dfrac{x+3}{x-1}}+3\sqrt{\dfrac{x-1}{x+3}}=4\\ \Leftrightarrow\dfrac{x+3}{x-1}+6\sqrt{\dfrac{x+3}{x-1}.\dfrac{x-1}{x+3}}+9.\dfrac{x-1}{x+3}=16\\ \Leftrightarrow\dfrac{\left(x+3\right)^2+9\left(x-1\right)^2}{\left(x+3\right)\left(x-1\right)}=10\\ \Leftrightarrow x^2+6x+9+9\left(x^2-2x+1\right)=10\left(x+3\right)\left(x-1\right)\\ \Leftrightarrow x^2+6x+9+9x^2-18x+9=10\left(x^2+3x-x-3\right)\\ \Leftrightarrow10x^2-12x+18=10x^2+20x-30\\ \Leftrightarrow32x-48=0\\ \Leftrightarrow x=\dfrac{3}{2}\)
\(\sqrt{\dfrac{x+3}{x-1}}+3\sqrt{\dfrac{x-1}{x+3}}=4\)
ĐK: \(x\in\left(-\infty;-3\right)\cup\left(1;\infty\right)\)
\(\Leftrightarrow\sqrt{\dfrac{x+3}{x-1}}-3+3\sqrt{\dfrac{x-1}{x+3}}-1=0\)
\(\Leftrightarrow\dfrac{\dfrac{8x-12}{1-x}}{\sqrt{\dfrac{x+3}{x-1}}+3}+3\cdot\dfrac{\dfrac{8x-12}{9x+27}}{\sqrt{\dfrac{x-1}{x+3}}+\dfrac{1}{3}}=0\)
\(\Leftrightarrow4\left(2x-3\right)\left(\dfrac{\dfrac{1}{1-x}}{\sqrt{\dfrac{x+3}{x-1}}+3}+\dfrac{\dfrac{3}{9x+27}}{\sqrt{\dfrac{x-1}{x+3}}+\dfrac{1}{3}}\right)=0\)
\(\Rightarrow2x-3=0\Rightarrow x=\dfrac{3}{2}\)
\(\sqrt{\dfrac{x+3}{x-1}}+3\sqrt{\dfrac{x-1}{x+3}}=4\left(1\right)\)
Đặt \(\sqrt{\dfrac{x+3}{x-1}}=a\left(a\ge0\right)\Rightarrow\sqrt{\dfrac{x-1}{x+3}}=\dfrac{1}{a}\)
\(\left(1\right)\Rightarrow a+\dfrac{3}{a}=4\)
\(\Leftrightarrow a^2+3=4a\)
\(\Leftrightarrow\left(a-1\right)\left(a-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=3\end{matrix}\right.\left(\text{nhận}\right)\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{\dfrac{x+3}{x-1}}=1\\\sqrt{\dfrac{x+3}{x-1}}=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x+3}{x-1}=1\\\dfrac{x+3}{x-1}=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=x-1\\x+3=9x-9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3=-1\left(\text{vô lý}\right)\\x=\dfrac{3}{2}\left(\text{nhận}\right)\end{matrix}\right.\)
Vậy (1) có tập no \(S=\left\{\dfrac{3}{2}\right\}\)
