ĐKXĐ:\(x\ne-1,x\ne0\)
\(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x+1}{x^2+x}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}-\dfrac{2x+1}{x\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-1+x-2x-1}{x\left(x+1\right)}=0\\ \Rightarrow x^2-x-2=0\\ \Leftrightarrow x^2-2x+x-2=0\\ \Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm `S={2}`
\(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x+1}{x^2+x}\left(đk:x\ne0,-1\right)\)
\(\Leftrightarrow\dfrac{x-1}{x}+\dfrac{1}{x+1}-\dfrac{2x+1}{x\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)+x-2x-1}{x\left(x+1\right)}=0\)
\(\Leftrightarrow x^2+x-x-1+x-2x-1=0\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Delta=b^2-4ac=\left(-1\right)^2-4.\left(-2\right)=9>0\Rightarrow\sqrt{\Delta}=3\)
\(\Rightarrow\)PT có 2 nghiệm \(x_1,x_2\)
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{1+3}{2}=2\left(n\right)\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{1-3}{2}=-1\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
