Câu 5:
\(a,\Leftrightarrow2019\left|x-2019\right|-2018\left|x-2019\right|+\left(x-2019\right)^2=0\\ \Leftrightarrow\left|x-2019\right|+\left(x-2019\right)^2=0\)
Lại có \(\left|x-2019\right|\ge0;\left(x-2019\right)^2\ge0\)
Do đó \(VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\left|x-2019\right|=0\\\left(x-2019\right)^2=0\end{matrix}\right.\Leftrightarrow x=2019\)
\(b,2x+\dfrac{1}{7}=\dfrac{1}{y}\\ \Leftrightarrow\dfrac{14x+1}{7}=\dfrac{1}{y}\\ \Leftrightarrow y\left(14x+1\right)=7=1.7\left(y\in Z^+\right)\)
Với \(\left\{{}\begin{matrix}y=1\\14x+1=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=\dfrac{4}{7}\left(loại\right)\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}y=7\\14x+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=7\\x=0\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(0;7\right)\)
