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\(x.y+2y+x=6\)
\(\Rightarrow y.\left(x+2\right)+\left(x+2\right)-2=6\)
\(\Rightarrow y.\left(x+2\right)+\left(x+2\right)=8\)
\(\Rightarrow\left(x+2\right).\left(y+1\right)=8\)
\(\Rightarrow\left(x+2\right).\left(y+1\right)\inƯ\left(8\right)=\left\{1;2;4;8\right\}\) mà : \(x+2\ge2\)
\(\Rightarrow\) \(x+2=2\Rightarrow x=0\)
\(y+1=4\Rightarrow y=3\)
\(\Rightarrow x=0;y=3\)
x.(y+3)-y=-2
\(\Rightarrow\)x.( y + 3 ) = y - 2
\(\Rightarrow\)xy + 3x = y - 2
\(\Rightarrow\) y( x - 1 ) + 3( x - 1 ) + 5 = 0
\(\Rightarrow\)y( x - 1 ) + 3( x - 1 ) = -5
\(\Rightarrow\)( y + 3 )( x - 1 ) = -5
\(\Rightarrow\)( y + 3 )( x - 1 ) \(\in\)Ư(-5) = { \(\pm1;\pm5\)}
Ta có bảng sau :
| y + 3 | - 1 | 5 | 1 | -5 |
| x - 1 | 1 | - 5 | 5 | -1 |
| x | 2 | - 4 | - 2 | - 8 |
| y | - 4 | 2 | 6 | 0 |
Vì \(\left(x,y\right)=5\) nên ta có: \(\hept{\begin{cases}x⋮5\\y⋮5\end{cases}}\Rightarrow\hept{\begin{cases}x=5m\\y=5n\\\left(m,n\right)=1\end{cases}}\)
Mà \(xy=825\)
\(\Rightarrow5m.5n=825\)
\(\Rightarrow25m.n=825\)
\(\Rightarrow mn=33\)
\(\left(m,n\right)=1\), ta có bảng sau:
| m | 1 | 33 | 3 | 11 |
| n | 33 | 1 | 11 | 3 |
| x | 5 | 165 | 15 | 55 |
| y | 165 | 5 | 55 | 15 |
Vậy \(\left(x;y\right)\in\left\{\left(5;165\right);\left(165;5\right)\left(15;55\right);\left(55;15\right)\right\}\).
x=1; y=3
em ghi sai đề ạ
\(x^2y+xy=6\)
=>\(y\left(x^2+1\right)=6\)
=>\(\left(x^2+1\right)\cdot y=1\cdot6=2\cdot3=3\cdot2=6\cdot1\)
=>\(\left(x^2+1;y\right)\in\left\{\left(1;6\right);\left(2;3\right);\left(3;2\right);\left(6;1\right)\right\}\)
=>\(\left(x^2;y\right)\in\left\{\left(0;6\right);\left(1;3\right);\left(2;2\right);\left(5;1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(0;6\right);\left(1;3\right);\left(-1;3\right);\left(\sqrt{2};2\right);\left(-\sqrt{2};2\right);\left(\sqrt{5};1\right);\left(-\sqrt{5};1\right)\right\}\)
mà (x,y) nguyên
nên \(\left(x;y\right)\in\left\{\left(0;6\right);\left(1;3\right);\left(-1;3\right)\right\}\)