ĐKXĐ: x>=-3
\(x^2+2x+7=3\sqrt{\left(x^2+1\right)\left(x+3\right)}\)
=>\(x^2+1-3\cdot\sqrt{\left(x^2+1\right)\left(x+3\right)}+2x+6=0\)
=>\(\left(\sqrt{x^2+1}\right)^2-\sqrt{\left(x^2+1\right)\left(x+3\right)}-2\cdot\sqrt{\left(x^2+1\right)\left(x+3\right)}+2\left(x+3\right)=0\)
=>\(\sqrt{x^2+1}\left(\sqrt{x^2+1}-\sqrt{x+3}\right)-2\cdot\sqrt{x+3}\left(\sqrt{x^2+1}-\sqrt{x+3}\right)=0\)
=>\(\left(\sqrt{x^2+1}-2\sqrt{x+3}\right)\left(\sqrt{x^2+1}-\sqrt{x+3}\right)=0\)
TH1: \(\sqrt{x^2+1}-2\sqrt{x+3}=0\)
=>\(\sqrt{x^2+1}=2\sqrt{x+3}=\sqrt{4x+12}\)
=>\(x^2+1=4x+12\)
=>\(x^2-4x-11=0\)
=>\(x^2-4x+4-15=0\)
=>\(\left(x-2\right)^2=15\)
=>\(\left[\begin{array}{l}x-2=\sqrt{15}\\ x-2=-\sqrt{15}\end{array}\right.=>\left[\begin{array}{l}x=\sqrt{15}+2\left(nhận\right)\\ x=-\sqrt{15}+2\left(nhận\right)\end{array}\right.\)
TH2: \(\sqrt{x^2+1}-\sqrt{x+3}=0\)
=>\(\sqrt{x^2+1}=\sqrt{x+3}\)
=>\(x^2+1=x+3\)
=>\(x^2-x-2=0\)
=>\(\left(x-2\right)\left(x+1\right)=0\)
=>x=2(nhận) hoặc x=-1(nhận)
