ĐKXĐ: \(x\ge0\)
Ta có: \(\left(x+4\right)^2-6\sqrt{x^3+3x}=13\) \(\Rightarrow x^2+8x+16-6\sqrt{x\left(x^2+3\right)}-13=0\)
\(\Rightarrow x^2+8x+3-6\sqrt{x\left(x^2+3\right)}=0\)\(\Rightarrow8x+\left(x^2+3\right)-6\sqrt{x\left(x^2+3\right)}=0\)
Chia 2 vế cho x2 + 3 ta được pt:
\(\Rightarrow8\frac{x}{x^2+3}+1-6\sqrt{\frac{x}{x^2+3}}=0\)
Đặt \(a=\sqrt{\frac{x}{x^2+3}}\left(a\ge0\right)\) ta được: 8a2 + 1 - 6a = 0 => a = 1/2 hoặc a = 1/4
+) Với a = 1/2 \(\Rightarrow\sqrt{\frac{x}{x^2+3}}=\frac{1}{2}\Rightarrow\frac{x}{x^2+3}=\frac{1}{4}\Rightarrow x^2+3=4x\Rightarrow x^2-4x+3=0\Rightarrow\left[\begin{array}{nghiempt}x=3\left(n\right)\\x=1\left(n\right)\end{array}\right.\)
+) Với a = 1/4 \(\Rightarrow\sqrt{\frac{x}{x^2+3}}=\frac{1}{4}\Rightarrow\frac{x}{x^2+3}=\frac{1}{8}\Rightarrow x^2+3=8x\Rightarrow x^2-8x+3=0\Rightarrow\left[\begin{array}{nghiempt}x=4+\sqrt{13}\left(n\right)\\x=4-\sqrt{13}\left(n\right)\end{array}\right.\)
Vậy \(S=\left\{1;3;4+\sqrt{13};4-\sqrt{13}\right\}\)
