Đk : \(x\ge0\)
\(\sqrt{x+1}+\sqrt{x+8}=3x+\sqrt{2x}\Leftrightarrow3x-\sqrt{x+8}+\sqrt{2x}-\sqrt{x+1}=0\Leftrightarrow\dfrac{\left(3x-\sqrt{x+8}\right)\left(3x+\sqrt{x+8}\right)}{3x+\sqrt{x+8}}+\dfrac{\left(\sqrt{2x}-\sqrt{x+1}\right)\left(\sqrt{2x}+\sqrt{x+1}\right)}{\sqrt{2x}+\sqrt{x+1}}=0\Leftrightarrow\dfrac{9x^2-x-8}{3x+\sqrt{x+8}}+\dfrac{x-1}{\sqrt{2x}+\sqrt{x+1}}=0\)
\(\dfrac{\left(x-1\right)\left(9x+8\right)}{3x+\sqrt{x+8}}+\dfrac{x-1}{\sqrt{2x}+\sqrt{x+1}}=0\Leftrightarrow\left(x-1\right)\left(\dfrac{\cdot9x+8}{3x+\sqrt{x+8}}+\dfrac{1}{\sqrt{2x}+\sqrt{x+1}}\right)=0\left(1\right)\)
Ta thấy: \(\dfrac{9x+8}{3x+\sqrt{x+8}}+\dfrac{1}{\sqrt{2x}+\sqrt{x+1}}>0\) ( với x \(\ge\) 0). Nên:
(1) \(\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\)
Vậy nghiệm của hệ phương trình là x = 1