- ĐKXĐ : \(\left\{{}\begin{matrix}2x-5\ge0\\x+2+3\sqrt{2x-5}\ge0\\x-2-\sqrt{2x-5}\ge0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ge\frac{5}{2}\\x+2+3\sqrt{2x-5}\ge0\left(TM\right)\\x-2\ge\sqrt{2x-5}\end{matrix}\right.\)
=> \(x\ge\frac{5}{2}\)
- Đặt \(\sqrt{2x-5}=a\left(a\ge0\right)\)
=> \(a^2=2x-5\)
=> \(x=\frac{a^2+5}{2}\)
- Thay x vào phương trình trên ta được :
\(\sqrt{\frac{a^2+5}{2}+2+3a}+\sqrt{\frac{a^2+5}{2}-2-a}=2\sqrt{2}\)
=> \(\sqrt{\frac{a^2+5+4+6a}{2}}+\sqrt{\frac{a^2+5-4-2a}{2}}=2\sqrt{2}\)
=> \(\sqrt{\frac{a^2+6a+9}{2}}+\sqrt{\frac{a^2-2a+1}{2}}=2\sqrt{2}\)
=> \(\frac{\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-1\right)^2}}{\sqrt{2}}=2\sqrt{2}\)
=> \(\left|a+3\right|+\left|a-1\right|=4\)
=> \(a+3+\left|a-1\right|=4\)
=> \(\left|a-1\right|=1-a\)
=> \(\left[{}\begin{matrix}a-1=a-1\\a-1=1-a\end{matrix}\right.\)
=> \(2a=2\)
=> \(a=1\) ( TM )
- Thay lại x vào phương trình trên ta được :
\(\sqrt{2x-5}=1\)
=> \(2x-5=1\)
=> \(x=3\) ( TM )
Vậy ....
