ĐKXĐ:\(x\ge\dfrac{3}{2}\)
\(x\sqrt{2x-3}=3x-4\\ \Leftrightarrow x^2\left(2x-3\right)=\left(3x-4\right)^2\\ \Leftrightarrow2x^3-3x^2=9x^2-24x+16\\ \Leftrightarrow2x^3-12x^2+24x-16=0\\ \Leftrightarrow x^3-6x^2+12x-8=0\\ \Leftrightarrow\left(x-2\right)^3=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\left(tm\right)\)