\(\left\{{}\begin{matrix}\sqrt{x+y\left(x-1\right)}+\sqrt{x}=y+\sqrt{y}\left(1\right)\\\left(x-1\right)^2+y\sqrt{\left(x-\dfrac{1}{y}\right)^3}=2\left(2\right)\end{matrix}\right.\)
\(\Rightarrow\left(1\right)\Leftrightarrow\sqrt{x+y\left(x-1\right)}-y-\sqrt{y}+\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{x+xy-y-y^2}{\sqrt{x+y\left(x-1\right)}+y}+\dfrac{x-y}{\sqrt{x}+\sqrt{y}}=0\)
\(\Leftrightarrow\dfrac{\left(x-y\right)\left(y+1\right)}{\sqrt{x+y\left(x-1\right)}+y}+\dfrac{x-y}{\sqrt{x}+\sqrt{y}}=0\)
\(\Leftrightarrow\left(x-y\right)\left(\dfrac{y+1}{\sqrt{x+y\left(x-1\right)}+y}+\dfrac{1}{\sqrt{x}+\sqrt{y}}\right)=0\)
\(\Leftrightarrow x=y\)
Thế vô (2) ta được
\(\left(2\right)\Leftrightarrow\left(x-1\right)^2+x\sqrt{\left(x-\dfrac{1}{x}\right)^3}=2\)
\(\Leftrightarrow x\sqrt{\left(x-\dfrac{1}{x}\right)^3}=2-\left(x-1\right)^2\)
\(\Leftrightarrow x^6-x^5+x^4-2x^3-x^2-x-1=0\)
\(\Leftrightarrow\left(x^2+1\right)^2\left(x^2-x-1\right)=0\)
\(\Leftrightarrow x^2-x-1=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y=\dfrac{1+\sqrt{5}}{2}\\x=y=\dfrac{1-\sqrt{5}}{2}\left(l\right)\end{matrix}\right.\)
