a/ \(mx^2-4x-3m+6=0\)
Để pt có nghiệm duy nhất
\(\Rightarrow\left[{}\begin{matrix}m=0\\\Delta'=4-m\left(-3m+6\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=0\\3m^2-6m+4=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow m=0\)
b/ \(\left[{}\begin{matrix}m=0\\\Delta'=\left(m+1\right)^2-m\left(m+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=0\\m=-1\end{matrix}\right.\)
c/ \(2x^2-2=mx^2+x\Leftrightarrow\left(m-2\right)x^2+x+2=0\)
Để pt có nghiệm duy nhất
\(\Rightarrow\left[{}\begin{matrix}m-2=0\\\Delta=1-8\left(m-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=2\\m=\frac{17}{8}\end{matrix}\right.\)
