$D=[\frac{-1}{2} ; \frac{3}{2}]$Bài 3: gpt : $\sqrt{2x+1}+\sqrt{3-2x}=\frac{1}{2}(2x-1)^2$
$-> 2\sqrt{2x+1} -(2x+1) + 2\sqrt{3-2x} -(3-2x) = 4x^2-4x-3$
$-> \frac{(2x+1)(3-2x)}{2\sqrt{2x+1} + (2x+1)} +\frac{(2x+1)(3-2x)}{2\sqrt{3-2x} + (3-2x)} = (2x+1)(2x-3)$
$-> (2x+1)(3-2x).[\frac{1}{2\sqrt{2x+1} + (2x+1)} + \frac{1}{2\sqrt{3-2x} + (3-2x)} + 1=0$
$-> x= -\frac{1}{2} v x=\frac{3}{2}$ (so với đk ta dễ thấy vế sau vô nghiệm)