\(x-y+2xy=7\)
\(\Rightarrow2x-2y+4xy=14\)
\(\Rightarrow2x\left(1+2y\right)-\left(2y+1\right)=13\)
\(\Rightarrow\left(2y+1\right)\left(2x-1\right)=13\)
làm nốt
do \(7\left(x-2004\right)^2\ge0\)\(\Rightarrow23-y^2\ge0\)
\(\Rightarrow y^2\le23\)
\(\Rightarrow y^2\in\left\{16,9,4,1,0\right\}\)
\(\Rightarrow y\in\left\{\pm1,\pm2,\pm3,\pm4\right\}\)
thay vào rồi tìm x
đề thiếu x,y thuộc Z
\(a,x-y+2xy=7\)
\(\Rightarrow x\left(2y-1\right)-y=7\)
\(\Rightarrow2x\left(2y-1\right)-2y=14\)
\(\Rightarrow2x\left(2y+1\right)-\left(2y+1\right)=14-1\)
\(\Rightarrow\left(2x-1\right)\left(2y+1\right)=13\)
\(\Rightarrow\left(2x-1\right)và\left(2y+1\right)\inƯ\left(13\right)=\left(\pm1,\pm13\right)\)
\(TH1:\hept{\begin{cases}2x-1=1\\2y+1=13\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=6\end{cases}}}\) \(TH2:\hept{\begin{cases}2x-1=-1\\2y+1=-13\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=-7\end{cases}}}\)
\(TH3:\hept{\begin{cases}2x-1=13\\2y+1=1\end{cases}\Rightarrow\hept{\begin{cases}x=7\\y=0\end{cases}}}\) \(TH4:\hept{\begin{cases}2x-1=-13\\2y+1=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-6\\y=-1\end{cases}}}\)
Vậy \(\left(x;y\right)\in\left\{\left(1;6\right);\left(0;-7\right);\left(7;0\right);\left(-6;-1\right)\right\}\)
\(b,7\left(x-2004\right)^2=23-y^2\)
\(\Rightarrow23-y^2\ge0\Leftrightarrow0\le y^2\le23\)
\(\Rightarrow y^2\in\left\{0;1;4;9;16\right\}\)
\(7\left(x-2004\right)^2⋮7\Rightarrow23-y^2⋮7\Rightarrow y^2\in\left\{9;16\right\}\)
+ Nếu \(y^2=9\Rightarrow y=3\Rightarrow7\left(x-2004\right)^2=23-9\)
\(\Rightarrow\left(x-2004\right)^2=2\left(loại\right)\)
+Nếu \(y^2=16\Rightarrow y=4\Rightarrow7\left(x-2004\right)^2=23-16\)
\(\Rightarrow\left(x-2004\right)^2=1=\left(\pm1\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-2004=1\\x-2004=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=2005\\x=-2003\end{cases}}}\)
Vậy x=2005 và y =1
x=-2003 và y =1
