ĐKXĐ: x>=-1/3
Ta có: \(\left(\sqrt{3x+1}-\sqrt{x+2}\right)\left(\sqrt{3x^2+7x+2}+4\right)=4x-2\)
=>\(\left(\sqrt{3x+1}-\sqrt{x+2}\right)\left(\sqrt{3x+1}+\sqrt{x+2}\right)\left(\sqrt{3x^2+7x+2}+4\right)=\left(4x-2\right)\left(\sqrt{3x+1}+\sqrt{x+2}\right)\)
=>\(\left(2x-1\right)\cdot\left(\sqrt{3x^2+7x+2}+4\right)=2\left(2x-1\right)\left(\sqrt{3x+1}+\sqrt{x+2}\right)\)
=>\(\sqrt{3x^2+7x+2}+4=2\cdot\left(\sqrt{3x+1}+\sqrt{x+2}\right)\)
=>\(\sqrt{\left(3x+1\right)\left(x+2\right)}-2\cdot\sqrt{3x+1}-2\cdot\sqrt{x+2}+4=0\)
=>\(\sqrt{3x+1}\left(\sqrt{x+2}-2\right)-2\left(\sqrt{x+2}-2\right)=0\)
=>\(\left(\sqrt{3x+1}-2\right)\left(\sqrt{x+2}-2\right)=0\)
TH1: \(\sqrt{3x+1}-2=0\)
=>\(\sqrt{3x+1}=2\)
=>3x+1=4
=>3x=3
=>x=1(nhận)
TH2: \(\sqrt{x+2}-2=0\)
=>\(\sqrt{x+2}=2\)
=>x+2=4
=>x=2(nhận)
