Đặt \(\sqrt{2x^2+x+9}=a>0\\ \sqrt{2x^2-x+1}=b>0\)
Ta có: \(\left\{{}\begin{matrix}a+b=x+4\\\dfrac{a^2-b^2}{2}=x+4\end{matrix}\right.\)
\(Pt\Leftrightarrow a+b=\dfrac{a^2-b^2}{2}\\ \Leftrightarrow\left(a-b\right)\left(a+b\right)-2\left(a+b\right)=0\\ \Leftrightarrow\left(a+b\right)\left(a-b-2\right)=0\\ \Leftrightarrow a-b-2=0\left(do.a+b>0\right)\\ \Leftrightarrow a=b+2\\ \Leftrightarrow\sqrt{2x^2+x+9}=\sqrt{2x^2-x+1}+2\\ \Leftrightarrow2x^2+x+9=2x^2-x+1+4+4\sqrt{2x^2-x+1}\)
\(\Leftrightarrow x+2=2\sqrt{2x^2-x+1}\left(x\ge-2\right)\\ \Leftrightarrow x^2+4x+4=8x^2-4x+4\\ \Leftrightarrow7x^2-8x=0\\ \Leftrightarrow x\left(7x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{8}{7}\left(tm\right)\end{matrix}\right.\)