\(\left|2x-5\right|+\left|xy-3y+2\right|=0\)
Ta có:
\(\left\{{}\begin{matrix}\left|2x-5\right|\ge0\\\left|xy-3y+2\right|\ge0\end{matrix}\right.\forall x,y.\)
\(\Rightarrow\left|2x-5\right|+\left|xy-3y+2\right|\ge0\) \(\forall x,y.\)
\(\Rightarrow\left|2x-5\right|+\left|xy-3y+2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|xy-3y+2\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-5=0\\xy-3y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=5\\xy-3y+2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5:2\\xy-3y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\\frac{5}{2}y-3y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\-\frac{1}{2}y+2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\-\frac{1}{2}y=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=\left(-2\right):\left(-\frac{1}{2}\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=4\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};4\right\}.\)
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