ĐKXĐ: \(\begin{cases}2x+1\ge0\\ 3-2x\ge0\end{cases}\Rightarrow\begin{cases}2x\ge-1\\ 2x\le3\end{cases}\Rightarrow-\frac12\le x\le\frac32\)
\(\sqrt{2x+1}+\sqrt{3-2x}=\frac{\left(2x-1\right)^2}{2}\)
=>\(\sqrt{2x+1}+\sqrt{3-2x}-2=\frac{\left(2x-1\right)^2}{2}-2=\frac{\left(2x-1\right)^2-4}{2}\)
=>\(\sqrt{2x+1}+\frac{3-2x-4}{\sqrt{3-2x}+2}=\frac{\left(2x-1-2\right)\left(2x-1+2\right)}{2}\)
=>\(\frac{2x+1}{\sqrt{2x+1}}-\frac{2x+1}{\sqrt{3-2x}+2}=\frac{\left(2x-3\right)\left(2x+1\right)}{2}\)
=>\(\left(2x+1\right)\left(\frac{1}{\sqrt{2x+1}}-\frac{1}{\sqrt{3-2x}+2}-\frac{2x-3}{2}\right)=0\)
TH1: 2x+1=0
=>2x=-1
=>\(x=-\frac12\) (nhận)
TH2: \(\frac{1}{\sqrt{2x+1}}-\frac{1}{\sqrt{3-2x}+2}-\frac{2x-3}{2}=0\)
=>\(\frac{1}{\sqrt{2x+1}}-\frac12+\frac12-\frac{1}{\sqrt{3-2x}+2}-\frac{2x-3}{2}=0\)
=>\(\frac{2-\sqrt{2x+1}}{2\cdot\sqrt{2x+1}}+\frac{\sqrt{3-2x}+2-2}{2\cdot\left(\sqrt{3-2x}+2\right)}-\frac{2x-3}{2}=0\)
=>\(\frac{4-2x-1}{2\cdot\sqrt{2x+1}\left(2+\sqrt{2x+1}\right)}+\frac{\sqrt{3-2x}}{2\left(\sqrt{3-2x}+2\right)}-\frac{2x-3}{2}=0\)
=>\(-\frac{2x-3}{2\cdot\sqrt{2x+1}\left(2+\sqrt{2x+1}\right)}+\frac{3-2x}{2\sqrt{3-2x}\left(\sqrt{3-2x}+2\right)}-\frac{2x-3}{2}=0\)
=>\(\left(2x-3\right)\left(-\frac{1}{2\sqrt{2x+1}\left(2+\sqrt{2x+1}\right)}-\frac{1}{2\cdot\sqrt{3-2x}\left(\sqrt{3-2x}+2\right)}-\frac12\right)\) =0
=>2x-3=0
=>2x=3
=>\(x=\frac32\) (nhận)
