$x^{2}$ + 5 = 2$\sqrt[]{2x+3}$ - 4x ĐKXĐ: x $\geq$ $\frac{-3}{2}$
(=) $x^{2}$ + 5 - 2$\sqrt[]{2x+3}$ + 4x = 0
(=) ( 2x +3 - 2 $\sqrt[]{2x+3}$ +1) + ($\sqrt[]{}$ $x^{2}$ +2x+1) = 0
(=) ($\sqrt[]{2x+3}$ - 1)$^{2}$ + (x + 1)$^{2}$ = 0
vì ($\sqrt[]{2x+3}$ - 1)$^{2}$ $\geq$ 0 với mọi x
(x + 1)$^{2}$ $\geq$ 0 với mọi x
mà ($\sqrt[]{2x+3}$ - 1)$^{2}$ + (x + 1)$^{2}$ = 0
=> $\left \{ {{(\sqrt[]{2x+3} - 1)^{2}=0} \atop {(x + 1)^{2}=0}} \right.$
=> $\left \{ {{\sqrt[]{2x+3} - 1=0} \atop {x + 1=0}} \right.$
=> x= -1 (tm)
S={-1}
