\(ĐK:x\ge0\\ 2\sqrt{x}+\sqrt{3x+2}=2+\sqrt{x+4}\\ \Leftrightarrow2\left(\sqrt{x}-1\right)+\left(\sqrt{3x+2}-\sqrt{5}\right)=\left(\sqrt{x+4}-\sqrt{5}\right)\\ \Leftrightarrow\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\left(\sqrt{3x+2}-\sqrt{5}\right)\left(\sqrt{3x+2}+\sqrt{5}\right)}{\left(\sqrt{3x+2}+\sqrt{5}\right)}-\dfrac{\left(\sqrt{x+4}-\sqrt{5}\right)\left(\sqrt{x+4}+\sqrt{5}\right)}{\left(\sqrt{x+4}+\sqrt{5}\right)}=0\\ \Leftrightarrow\dfrac{2\left(x-1\right)}{\sqrt{x}+1}+\dfrac{3\left(x-1\right)}{\sqrt{3x+2}+\sqrt{5}}-\dfrac{x-1}{\sqrt{x+4}+\sqrt{5}}=0\\ \)\(\Leftrightarrow\left(x-1\right)\left(\dfrac{2}{\sqrt{x}+1}+\dfrac{3}{\sqrt{3x+2}+\sqrt{5}}-\dfrac{1}{\sqrt{x+4}+\sqrt{5}}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\\dfrac{2}{\sqrt{x}+1}+\dfrac{3}{\sqrt{3x+2}+\sqrt{5}}-\dfrac{1}{\sqrt{x+4}+\sqrt{5}}=0\end{matrix}\right.\)
\(\dfrac{2}{\sqrt{x}+1}+\dfrac{3}{\sqrt{3x+2}+\sqrt{5}}-\dfrac{1}{\sqrt{x+4}+\sqrt{5}}>0\)
Vậy x=1
(vì
