ĐKXĐ : \(\left\{{}\begin{matrix}x\ne-1\\x\ne0\end{matrix}\right.\)
\(\dfrac{2x-1}{x\left(x+1\right)}=\dfrac{1}{x}\)
\(\Leftrightarrow\) \(\dfrac{x\left(2x-1\right)}{x^2\left(x+1\right)}=\dfrac{x\left(x+1\right)}{x^2\left(x+1\right)}\)
\(\Leftrightarrow x\left(2x-1\right)=x\left(x+1\right)\)
\(\Leftrightarrow2x^2-x-x^2-x=0\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=2\left(N\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
Đúng 0
Bình luận (0)