a) ĐKXĐ: \(x\ge-\frac{3}{2}\)
Ta có: \(\sqrt{x^2+4}=\sqrt{2x+3}\)
\(\Leftrightarrow\left(\sqrt{x^2+4}\right)^2=\left(\sqrt{2x+3}\right)^2\)
\(\Leftrightarrow x^2+4=2x+3\)
\(\Leftrightarrow x^2+4-2x-3=0\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)(nhận)
Vậy: x=1
b) ĐKXĐ: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{x^2-6x+9}=2x-1\)
\(\Leftrightarrow x^2-6x+9=\left(2x-1\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(x-3-2x+1\right)\left(x-3+2x-1\right)=0\)
\(\Leftrightarrow\left(-x-2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-2=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=2\\3x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=\frac{4}{3}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x=\frac{4}{3}\)
