Question

cho x,y nguyên dương thỏa mản (4x-1).(y-3)=18

tìm x,y

Answer 1

\(\left(4x-1\right)\left(y-3\right)=18\)

\(\Rightarrow\left(4x-1\right);\left(y-3\right)\in U\left(18\right)=\left\{1;2;3;6;9;18\right\}\left(x,y\inℤ^+\right)\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(\dfrac{1}{2};31\right);\left(\dfrac{3}{4};12\right);\left(1;9\right);\left(\dfrac{7}{4};6\right);\left(\dfrac{5}{2};5\right);\left(\dfrac{19}{4};4\right)\right\}\left(x,y\inℤ^+\right)\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;9\right)\right\}\left(x,y\inℤ^+\right)\)