\(\sqrt{x-\dfrac{1}{x}}-\sqrt{1-\dfrac{1}{x}}=\dfrac{x-1}{x}\)
\(\Leftrightarrow\sqrt{\dfrac{x^2-1}{x}}-\sqrt{\dfrac{x-1}{x}}=\dfrac{x-1}{x}\)
\(\Leftrightarrow\sqrt{x^3-x}-\sqrt{x^2-x}=x-1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^3-x}=a\\\sqrt{x^2-x}=b\end{matrix}\right.\left(a,b\ge0\right)\), ta có:
\(\left\{{}\begin{matrix}a^2-b^2=x^3-x^2\\a-b=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=x^2\left(x-1\right)\\a-b=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=x^2\\a-b=x-1\end{matrix}\right.\)
\(\Rightarrow b=\dfrac{x^2-x+1}{2}=\dfrac{b^2+1}{2}\)
\(\Rightarrow b^2+1=2b\)
\(\Leftrightarrow\left(b-1\right)^2=0\)
\(\Leftrightarrow b=1\)
\(\Rightarrow\sqrt{x^2-x}=1\)
\(\Leftrightarrow x^2-x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\left(n\right)\\x=\dfrac{1-\sqrt{5}}{2}\left(l\right)\end{matrix}\right.\)
Vậy . . . ^3^
