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a: \(\Leftrightarrow\left(x,y-3\right)\in\left\{\left(1;15\right);\left(3;5\right);\left(5;3\right);\left(15;1\right);\left(-1;-15\right);\left(-3;-5\right);\left(-5;-3\right);\left(-15;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(1;18\right);\left(3;8\right);\left(5;6\right);\left(15;4\right);\left(-1;-12\right);\left(-3;-2\right);\left(-5;0\right);\left(-15;2\right)\right\}\)
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\(xy-2y+3\left(x-2\right)=7\)
\(xy-2y+3x-3\cdot2=7\)
\(\left(x-2\right)y+\left(x-2\right)3=7\)
\(\left(x-2\right)\left(y+3\right)=7\)
Ta có: 1 x 7 = 7; 7 x1 = 7
(-1) x (-7) = 7 ; (-7) x (-1)= 7
=> TH1: x-2= 1 => x=2
y+3 = 7 => y= 4
TH2: x-2= 7 => x=9
y+3 = 1 => y= -2
TH3: x-2= -1 => x=1
y+3 = -7 => y= -10
TH4: x-2= -7 => x=-5
y+3 = -1 => y= -4
Vậy \(x\in\left\{-5;1;2;9\right\}\)
\(y\in\left\{-10;-4;-2;4\right\}\)
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Bài 1:
Để E nguyên thì \(x+5⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{3;1;9;-5\right\}\)
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xy -x - y =2
x.( y-1) = 2+ y
\(x=\frac{2+y}{y-1}=\frac{y-1+3}{y-1}=\frac{y-1}{y-1}+\frac{3}{y-1}=1+\frac{3}{y-1}\)
Để x nguyên
\(\Rightarrow\frac{3}{y-1}\in z\Rightarrow3⋮y-1\Rightarrow y-1\inƯ_{\left(3\right)}=\left(3;-3;1;-1\right)\)
nếu y- 1 = 3 => y =4 (TM) => x = 1+ 3/4-1 = 1 + 1 =2 => x= 2 (TM)
y-1 =-3 => y =-2 (TM) => x = 1+ 3/-2-1 = 1+(-1) =0 => x =0 (TM)
y -1 = 1 => y=2 (TM) => x = 1+ 3/2-1 = 1+3 =4 => x =4 (TM)
y-1 =-1 => y=0 (TM) => x = 1+ 3/0-1 = 1+(-3) = -2 => x = -1 (TM)
KL: (x;y) =........................................
Chúc bn học tốt !!!!!