a)Ta thấy: \(\left\{{}\begin{matrix}\left(12x-y+7\right)^{2016}\ge0\forall x,y\\\left|2x-3\right|^{2017}\ge0\forall x\end{matrix}\right.\)
\(\Rightarrow\left(12x-y+7\right)^{2016}+\left|2x-3\right|^{2017}\ge0\forall x,y\)
Mà \(\left(12x-y+7\right)^{2016}+\left|2x-3\right|^{2017}\le0\)
Nên xảy ra khi \(\left(12x-y+7\right)^{2016}+\left|2x-3\right|^{2017}=0\)
\(\left\{{}\begin{matrix}\left(12x-y+7\right)^{2016}=0\\\left|2x-3\right|^{2017}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}12x-y+7=0\\x=\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}y=25\\x=\dfrac{3}{2}\end{matrix}\right.\)
b)\(1+2+3+...+n=\overline{aaa}\)
Ta có: \(\left\{{}\begin{matrix}VT=\dfrac{n\left(n+1\right)}{2}\\VP=a\cdot111\end{matrix}\right.\)
\(\Rightarrow\dfrac{n\left(n+1\right)}{2}=a\cdot111\Rightarrow n\left(n+1\right)=a\cdot222\)
\(\Rightarrow n\left(n+1\right)=6a\cdot37=6a\left(36+1\right)\)
Dễ thấy: \(n\left(n+1\right)\) là \(2\) số tự nhiên liên tiếp và \(6a\) và \(36+1\) là 2 số tự nhiên liên tiếp
\(\Rightarrow6a=36\Rightarrow a=6\Rightarrow n=36\)
