\(x\sqrt{x}=x+\sqrt{x}\)
\(\Rightarrow\sqrt{x}\left(x-\sqrt{x}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\x-\sqrt{x}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\left(l\right)\\x-\sqrt{x}-1=0\left(1\right)\end{cases}}}\)
Từ (1) . Có: \(\Delta=\left(-1\right)^2-4\left(-1\right)=5>0\Rightarrow\sqrt{\Delta}=\sqrt{5}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{5}}{2}\left(n\right)\\x=\frac{1-\sqrt{5}}{2}\left(n\right)\end{cases}}\)
Vậy \(x=\left\{\frac{1+\sqrt{5}}{2};\frac{1-\sqrt{5}}{2}\right\}\)
đặt \(\sqrt{x}\)\(=y\)
=> \(x=y^2\)
ta có pt : \(y^3=\)\(y^2+y\)
\(\Leftrightarrow y^3-y^2-y=0\)
\(y\ne0\)\(\Rightarrow\)\(y^2-y-1=0\)
ptdttnt rồi bạn tự tính y => x
