\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\left(dkxd:x\ne0,x\ne-1\right)\)
\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}-\dfrac{2x-1}{x\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)-x-\left(2x-1\right)}{x\left(x+1\right)}=0\)
\(\Leftrightarrow x^2-1-x-2x+1=0\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=3\left(n\right)\end{matrix}\right.\)
Vậy \(S=\left\{3\right\}\)
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