\(a)\dfrac{x-1}{x+1}=\dfrac{1}{x-1}\left(DK:x\ne1;x\ne-1\right)\\ \Leftrightarrow\left(x-1\right)\left(x-1\right)=x+1\\ \Leftrightarrow\left(x-1\right)^2-x-1\\ \Leftrightarrow x^2-2x+1-x-1=0\\ \Leftrightarrow x^2-3x=0\\ \Leftrightarrow x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=3\left(TM\right)\end{matrix}\right.\)
Vậy S = {0; 3}
\(b)3x-5=7\\ \Leftrightarrow3x=12\\ \Leftrightarrow x=4\)
Vậy S = {4}
\(c)\dfrac{5}{x+3}=\dfrac{3}{x-1}\left(DK:x\ne-3;x\ne1\right)\\ \Leftrightarrow5\left(x-1\right)=3\left(x+3\right)\\ \Leftrightarrow5x-5=3x+9\\ \Leftrightarrow5x-3x=9+5\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\left(TM\right)\)
Vậy S = {7}
\(d)-2x+14=0\\ \Leftrightarrow-2x=-14\\ \Leftrightarrow x=7\)
Vậy S = {7}
\(e)2x\left(x-3\right)+5\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(S=\left\{3;-\dfrac{5}{2}\right\}\)
\(f)x^2-5x+6=0\\ \Leftrightarrow x^2-2x-3x+6=0\\ \Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy S = {2; 3}
