1/ \(\left(x+2y-3\right)^{2016}+\left|2x+3y-5\right|=0\)
Ta có: \(\left\{{}\begin{matrix}\left(x+2y-3\right)^{2016}\ge0\forall x,y\\\left|2x+3y-5\right|\ge0\forall x,y\end{matrix}\right.\)
Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+2y-3=0\\2x+3y-5=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+2y=3\\2x+3y=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;1\right)\)
2/ \(\dfrac{x+4}{x-1}< 0\)
TH1: \(\left\{{}\begin{matrix}x+4>0\\x+1< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>-4\\x< -1\end{matrix}\right.\)
\(\Rightarrow-4< x< -1\Rightarrow\left\{{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+4< 0\\x+1>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< -5\\x>-1\end{matrix}\right.\)(vô lý)
Vậy có 2 giá trị \(x\) t/m đó là:\(\left\{{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
