Tìm x,y ∈ Z biết:
\(3\left|2x+1\right|+4\left|2y-1\right|\le7\)
T đã hứa thì t sẽ làm:v
\(3\left|2x+1\right|+4\left|2y-1\right|\le7\)
\(\Rightarrow3\left|2x+1\right|\le7-4\left|2y-1\right|\le7\)
mà: \(\left\{{}\begin{matrix}3 \left|2x+1\right|\ge0\\3\left|2x+1\right|⋮3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\le3\left|2x+1\right|\le7\\3\left|2x+1\right|⋮3\end{matrix}\right.\)
Vì x nguyên nên: \(3\left|2x+1\right|\in\left\{0;3;6\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\left|2x+1\right|=0\\\left|2x+1\right|=1\\\left|2x+1\right|=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(loại\right)\\\left[{}\begin{matrix}2x+1=1\Leftrightarrow x=0\left(chọn\right)\\2x+1=-1\Leftrightarrow x=-1\left(chọn\right)\end{matrix}\right.\\\left[{}\begin{matrix}2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(loại\right)\\2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
Với \(\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) thì: \(3\left|2x+1\right|=3\Leftrightarrow4\left|2y-1\right|\le7-3=4\)
Vì \(y\in Z\) nên: \(\left[{}\begin{matrix}4\left|2y-1\right|=4\\4\left|2y-1\right|=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}2y-1=1\Leftrightarrow y=1\left(chọn\right)\\2y-1=-1\Leftrightarrow y=0\left(chọn\right)\end{matrix}\right.\\2y=1\Leftrightarrow y=\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
Vậy: \(\left(x;y\right)=\left(0;0\right);\left(0;1\right);\left(-1;1\right);\left(-1;0\right)\)
Dựa vào điều kiện: \(x;y\in Z\) là giải ra thôi bạn:v
Vì: \(x;y\in Z\) \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}0\le3\left|2x+1\right|\le7\\3\left|2x+1\right|⋮3\end{matrix}\right.\\\left\{{}\begin{matrix}0\le4\left|2y-1\right|\le7\\4\left|2y-1\right|⋮4\end{matrix}\right.\end{matrix}\right.\)
T chưa lm đc mi tề!
