ĐKXĐ: x∈R
Ta có: \(2x^2+3x+3=5\sqrt{2x^2+3x+9}\)
=>\(2x^2+3x+9-5\cdot\sqrt{2x^2+3x+9}-6=0\)
=>\(\left(\sqrt{2x^2+3x+9}-6\right)\left(\sqrt{2x^2+3x+9}+1\right)=0\)
=>\(\sqrt{2x^2+3x+9}-6=0\)
=>\(\sqrt{2x^2+3x+9}=6\)
=>\(2x^2+3x+9=6^2=36\)
=>\(2x^2+3x-27=0\)
=>\(2x^2+9x-6x-27=0\)
=>x(2x+9)-3(2x+9)=0
=>(2x+9)(x-3)=0
=>x=-9/2(nhận) hoặc x=3(nhận)