Đề bài <=> \(\sqrt{3\left(x^2+2x+1\right)+1}=3-\left(x^2+2x+1\right)\)
<=> \(\sqrt{3\left(x+1\right)^2+1}=3-\left(x+1\right)^2\)
Đặt \(\left(x+1\right)^2=a\left(a\ge0\right)\)
Có \(\sqrt{3a+1}=3-a\)
<=> \(3a+1=9-6a+a^2\)
<=> \(0=9-6a+a^2-3a-1\)
<=> \(8-9x+a^2=0\)
<=> \(\left(a-8\right)\left(a-1\right)=0\)
=> \(\left[{}\begin{matrix}a=8\\a=1\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\left(x+1\right)^2=8\\\left(x+1\right)^2=1\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x+1=2\sqrt{2}\\x+1=-2\sqrt{2}\\x+1=1\\x+1=-1\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\sqrt{2}-1\left(ktm\right)\\x=-2\sqrt{2}+1\left(ktm\right)\\x=0\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{0,-2\right\}\)
