ĐKXĐ : \(\left\{{}\begin{matrix}2x+9\ge0\\4-x\ge0\\3x+1\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}2x\ge-9\\-x\ge-4\\3x\ge-1\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x\ge-\frac{9}{2}\\x\le4\\x\ge-\frac{1}{3}\end{matrix}\right.\)
<=> \(4\ge x\ge-\frac{1}{3}\)
Ta có : \(\sqrt{2x+9}=\sqrt{4-x}+\sqrt{3x+1}\)
<=> \(\left(\sqrt{2x+9}\right)^2=\left(\sqrt{4-x}+\sqrt{3x+1}\right)^2\)
<=> \(2x+9=\left(4-x\right)+2\sqrt{\left(4-x\right)\left(3x+1\right)}+\left(3x+1\right)\)
<=> \(2x+9=4-x+2\sqrt{12x-3x^2+4-x}+3x+1\)
<=> \(2x+9-4+x-3x-1=2\sqrt{12x-3x^2+4-x}\)
<=> \(4=2\sqrt{12x-3x^2+4-x}\)
<=> \(4^2=\left(2\sqrt{12x-3x^2+4-x}\right)^2\)
<=> \(16=4\left(12x-3x^2+4-x\right)\)
<=> \(4=12x-3x^2+4-x\)
<=> \(0=12x-3x^2-x\)
<=> \(0=11x-3x^2\)
<=> \(0=x\left(11-3x\right)\)
<=> \(\left\{{}\begin{matrix}x=0\\11-3x=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\\-3x=-11\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\\x=\frac{11}{3}\end{matrix}\right.\) ( TM )