a ) \(2m-3m-6=0\)
\(\Leftrightarrow-m=6\)
\(\Leftrightarrow m=-6\)
Vậy ...
b ) \(\left|x-m\right|+\left|x^2+4x-5\right|=0\) (1)
Do \(\left\{{}\begin{matrix}\left|x-m\right|\ge0\\\left|x^2+4x-5\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-m\right|+\left|x^2+4x-5\right|\ge0\) (2)
Từ ( 1 ) ; ( 2 ) \(\Rightarrow\left\{{}\begin{matrix}x-m=0\\x^2+4x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m\\\left(x+5\right)\left(x-1\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m\\\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=-5\\m=1\end{matrix}\right.\)
Vậy ...
a, \(2x-3m-6=0\)
\(\Leftrightarrow2x=3m+6\)
\(\Leftrightarrow x=\dfrac{3m+6}{2}\)
Vậy với \(\forall m\) thì pt luôn có nghiệm .
b, \(\left|x-m\right|+\left|x^2+4x-5\right|=0\)
Do \(\left\{{}\begin{matrix}|x-m|\ge0\\\left|x^2+4x-5\right|\ge0\end{matrix}\right.\)
Suy ra :
\(\left\{{}\begin{matrix}x-m=0\\x^2+4x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m\\\left(x+5\right)\left(x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=-5\\m=1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}m=-5\\m=1\end{matrix}\right.\) thì pt có nghiệm
a, \(2m-3m-6=0\)
\(\Leftrightarrow-m=6\)
\(\Leftrightarrow m=-6\)
Vậy...
