\(\Leftrightarrow\left(2m-1\right)\left(2m+1\right)x-m\left(2m-1\right)=0\)
\(\Leftrightarrow\left(2m-1\right)\left(\left(2m+1\right)x-m\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2m-1=0\\\left(2m+1\right)x-m=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=\frac{1}{2}\left(1\right)\\x=\frac{m}{2m+1}\left(2\right)\end{matrix}\right.\)
b/Nếu (1) là nghiệm thì PT đúng \(\forall x\in R\) nên \(m\ne\frac{1}{2}\) để x có nghiệm duy nhất
Ta có (2)>1/2\(\Rightarrow\frac{m}{2m+1}>\frac{1}{2}\)\(\Leftrightarrow\frac{2m}{4m+2}-\frac{2m+1}{2m+2}>0\)
\(\Leftrightarrow-\frac{1}{4m+2}>0\)\(\Rightarrow4m+2< 0\Leftrightarrow2m+1< 0\Leftrightarrow m< -\frac{1}{2}\)