ĐKXĐ: \(x\ge\dfrac{1}{2}\)
Đặt \(\sqrt{2x-1}=y\ge0\)
\(\Rightarrow x^3+y^3+\dfrac{3xy}{x+y}=1\)
\(\Leftrightarrow\left(x+y\right)^3-1-3xy\left(x+y\right)+\dfrac{3xy}{x+y}=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1\right)-\dfrac{3xy\left(x+y-1\right)\left(x+y+1\right)}{x+y}=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-\dfrac{3xy\left(x+y+1\right)}{x+y}\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(\dfrac{x^3+y^3+\left(x^2-xy+y^2\right)+x+y}{x+y}\right)=0\)
\(\Leftrightarrow x+y-1=0\)
\(\Leftrightarrow y=1-x\)
\(\Leftrightarrow\sqrt{2x-1}=1-x\left(x\le1\right)\)
\(\Rightarrow2x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-4x+2=0\)
