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4:
(x+1)(y-2)=5
=>\(\left(x+1;y-2\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;7\right);\left(4;3\right);\left(-2;-3\right);\left(-6;1\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(x^2-xy+y+2=0\)
\(\Leftrightarrow\left(x-\dfrac{y}{2}\right)^2-\left(\dfrac{y}{2}-1\right)^2+3=0\)
\(\Leftrightarrow\left(x-\dfrac{y}{2}\right)^2-\left(\dfrac{y}{2}-1\right)^2=1-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{y}{2}\right)^2=1\\\left(\dfrac{y}{2}-1\right)^2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}2x-y=2\\2x-y=-2\end{matrix}\right.\\\left[{}\begin{matrix}y=6\\y=-2\end{matrix}\right.\end{matrix}\right.\)
với y=6 \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
với y=-2 \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
vậy S=\(\left\{\left(4;6\right);\left(2;6\right);\left(0;-2\right);\left(-2;-2\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-3x-\(\dfrac{2}{-20}=\dfrac{2}{5}\)
-3x+\(\dfrac{2}{20}=\dfrac{2}{5}\)
=>-3x=\(\dfrac{2}{5}-\dfrac{2}{20}\)
=>-3x=\(\dfrac{8-2}{20}\)
=>-3x=\(\dfrac{6}{20}=\dfrac{3}{10}\)
=>x=\(\dfrac{3}{10}.\dfrac{-1}{3}\)
=>x=\(\dfrac{-3}{30}=\dfrac{-1}{10}\)
Vậy x=.......
x còn =1 đúng k biết mình vẫn chưa tính