\(\dfrac{x+1}{x-1}-\dfrac{3x+1}{x^2-x}=\dfrac{1}{x}\left(x\ne1;x\ne0\right)\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{x\left(x-1\right)}-\dfrac{3x+1}{x\left(x-1\right)}=\dfrac{x-1}{x\left(x-1\right)}\)
\(\Leftrightarrow x\left(x+1\right)-\left(3x+1\right)=x-1\)
\(\Leftrightarrow x^2+x-3x-1=x-1\)
\(\Leftrightarrow x^2-2x-1=x-1\)
\(\Leftrightarrow x^2-2x-1-x+1=0\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\left(tm\right)\end{matrix}\right.\)
Vậy: ...
ĐK: \(\left\{{}\begin{matrix}x-1\ne0\\x^2-x\ne0\\x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne1\\x\ne0\end{matrix}\right.\)
PT tuong đương:
\(\dfrac{x\left(x+1\right)}{x\left(x-1\right)}-\dfrac{3x+1}{x\left(x-1\right)}=\dfrac{1\left(x-1\right)}{x\left(x-1\right)}\\ \Leftrightarrow x^2+x-3x-1-x+1=0\\ \Leftrightarrow x^2-3x=0\\ \Leftrightarrow x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
