\(a\text{) }\sqrt{2x-3}=2\sqrt{x-1}\left(đk:x\ge1\right)\\ \Leftrightarrow2x-3=4\left(x-1\right)\\ \Leftrightarrow2x-3=4x-4\\ \Leftrightarrow-2x=-1\\ \Leftrightarrow x=\dfrac{1}{2}\left(K^o\text{ }T/m\right)\)
\(\text{b) }\sqrt[3]{\dfrac{x+3}{6x-3}}=\sqrt{\dfrac{2}{3}}\\ \Leftrightarrow\left(\dfrac{x+3}{6x-3}\right)^2=\left(\dfrac{2}{3}\right)^3\\ \Leftrightarrow\dfrac{\left(x+3\right)^2}{9\left(2x-1\right)^2}=\dfrac{8}{27}\\ \Leftrightarrow\dfrac{x^2+6x+9}{4x^2-4x+1}=\dfrac{8}{3}\\ \Leftrightarrow3\left(x^2+6x+9\right)=8\left(4x^2-4x+1\right)\\ \Leftrightarrow3x^2+18x+27=32x^2-32x+8\\ \Leftrightarrow29x^2-50x-19=0\\ \Leftrightarrow x^2-\dfrac{50}{29}x-\dfrac{19}{29}=0\\ \Leftrightarrow x^2-\dfrac{50}{29}x+\dfrac{625}{841}-\dfrac{1176}{29}=0\\ \Leftrightarrow\left(x-\dfrac{25}{29}\right)^2-\dfrac{1176}{29}=0\\ \Leftrightarrow\left(x-\dfrac{25}{29}+\dfrac{14\sqrt{6}}{29}\right)\left(x-\dfrac{25}{29}-\dfrac{14\sqrt{6}}{29}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{25}{29}+\dfrac{14\sqrt{6}}{29}=0\\x-\dfrac{25}{29}-\dfrac{14\sqrt{6}}{29}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{25-14\sqrt{6}}{29}\\x=\dfrac{25+14\sqrt{6}}{29}\end{matrix}\right.\)