a: \(\Leftrightarrow6x^2+2x+8+\sqrt{3x^2+x+4}-18=0\)
\(\Leftrightarrow2\left(\sqrt[3]{3x^2+x+4}\right)^3+\sqrt[3]{3x^2+x+4}-18=0\)
=>\(3x^2+x+4=8\)
=>3x^2+x-4=0
=>x=1 hoặc x=-4/3
b: ĐKXĐ: x>0
Pt sẽ là \(x+8+9x-6\sqrt{x\left(x+8\right)}=0\)
=>\(10x+8=\sqrt{36x\left(x+8\right)}\)
=>36x^2+288x=100x^2+160x+64
=>x=1