1) a)\(A=\dfrac{1-2x}{x+3}=\dfrac{-2x+1}{x+3}\)
\(A\in Z\Rightarrow-2x+1⋮x+3\)
\(\Rightarrow-2x-6+7⋮x+3\)
\(\Rightarrow-2\left(x+3\right)+7⋮x+3\)
\(\Rightarrow7⋮x+3\)
\(\Rightarrow x+3\inƯ\left(7\right)\)
\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x+3=1\\x+3=-1\\x+3=7\\x+3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-4\\x=4\\x=-10\end{matrix}\right.\)
b)\(A=\dfrac{x+3}{x-2}=\dfrac{x-2+5}{x-2}=\dfrac{x-2}{x-2}+\dfrac{5}{x-2}=1+\dfrac{5}{x-2}\)
\(\Rightarrow5⋮x-2\Rightarrow x-2\inƯ\left(5\right)\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\\x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=7\\x=-3\end{matrix}\right.\)
2)
\(25-y^2=8\left(x-2009\right)^2\)
\(\left\{{}\begin{matrix}8\left(x-2009\right)^2\ge0\\8\left(x-2009\right)^2⋮8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}25-y^2\ge0\\25-y^2⋮8\end{matrix}\right.\)
Vậy \(0\le y^2\le25\)
\(\Leftrightarrow0\le y^2\le5^2\)
Vì \(y\in Z\) nên: \(y\in\left\{0;\pm1;\pm2;\pm3;\pm4;\pm5\right\}\)
\(y\in\left\{0;1;4;9;16;25\right\}\)
mà chỉ có :
\(25-25=0⋮8\Rightarrow y^2=25\Leftrightarrow y=\pm5\)
\(\Leftrightarrow8\left(x-2009\right)^2=0\Leftrightarrow x=2009\)
Vậy \(\left\{{}\begin{matrix}y=5\\x=2009\end{matrix}\right.\) và \(\left\{{}\begin{matrix}y=-5\\x=2009\end{matrix}\right.\)
