ĐKXĐ: \(x\ge-3\)
Đặt \(\sqrt{\frac{x+3}{2}}=a+1\ge0\Rightarrow x+3=2a^2+4a+2\)
Ta được hệ: \(\left\{{}\begin{matrix}2x^2+4x-a=1\\2a^2+4a-x=1\end{matrix}\right.\)
Trừ vế cho vế:
\(2\left(x^2-a^2\right)+4\left(x-a\right)+\left(x-a\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left(2x+2a+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=x\\2\left(a+1\right)=-2x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{\frac{x+3}{2}}=x+1\\2\sqrt{\frac{x+3}{2}}=-2x-3\end{matrix}\right.\) \(\Leftrightarrow...\)
ta có \(2x^2+4x=\sqrt{\frac{x+3}{2}}\)
⇔4x4+16x3+16x2=\(\frac{x+3}{2}\)
⇔x+3=8x4+32x3+32x2
⇔x+3-8x4-32x3-32x2=0
⇔10x-9x+3-8x4-12x3-20x3+4x2-30x2-6x2=0
⇔(-6x2-9x+3)+(-8x4-12x3+4x2)+(-20x3-30x2+10x)
⇔-3(2x2+3x-1)-4x2(2x2+3x-1)-10x(2x2+3x-1)
⇔-(2x2+3x-1)(4x2+10x+3)
⇔\(\left[{}\begin{matrix}2x^2+3x-1=0\\4x^2+10+3=0\end{matrix}\right.\)
1. 2x2+3x-1=0
⇔x2+\(\frac{3}{2}\)x-\(\frac{1}{2}\)=0
⇔(x+\(\frac{3}{4}\))2=\(\frac{17}{16}\)
⇔\(x=\left\{{}\begin{matrix}\frac{-3+\sqrt{17}}{4}\\\frac{-3-\sqrt{17}}{4}\end{matrix}\right.\)
2.tương tự
x= \(\left\{{}\begin{matrix}\frac{-5-\sqrt{13}}{4}\\\frac{-5+\sqrt{13}}{4}\end{matrix}\right.\)
thử lại nghiệm thì chỉ có \(\frac{-3+\sqrt{17}}{4}\) và\(\frac{-5-\sqrt{13}}{4}\)thỏa mãn
⇒x=\(\frac{-3+\sqrt{17}}{4}\) và x=\(\frac{-5-\sqrt{13}}{4}\)
hơi dài 
