tim x,y nguyen >0 tm a) 5(x+y)+2=3xy b)2(x+y)=5xy c)3x+7=y(x-3)

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1. a. `5(x+y)+2=3xy`

   `⇒5x-3xy+5y=-2`

   `⇒x(5-3y)+5y=-2`

   `⇒-3x(5-3y)-15y=6`

   `⇒-3x(5-3y)+(25-15y)=31`

   `⇒-3x(5-3y)+5(5-3y)=31`

   `⇒(5-3y)(5-3x)=31`

   `⇒(5-3y);(5-3x)∈Ư(31)={±1;±31}`

   `⇒(x;y)∈{(2;12);(4/3;-26/3);(12;2);(4/3;-26/3)}`

   Vì `x,y∈Z^+⇒(x;y)∈{(2;12);(12;2)}`

   b. `2(x+y)=5xy`

   `⇒2x-5xy+2y=0`

   `⇒x(2-5y)+2y=0`

   `⇒-5x(2-5y)+4-10y=4`

   `⇒(2-5y)(2-5x)=4`

   `⇒(2-5y);(2-5x)∈Ư(4)={±1;±2;±4}`

   Kết hợp với `x,y∈Z^+⇒(x;y)∈{(0;0)}`

   c. `3x+7=y(x-3)`

   `⇒(3x-9)-y(x-3)=-16`

   `⇒(x-3)(3-y)=-16`

   `⇒(x-3);(3-y)∈Ư(16)={±1;±2;±4;±8;±16}`

   Vì $\left \{ {{x>0} \atop {x∈Z}} \right.⇒\left \{ {{x-3>-3} \atop {x∈Z}} \right.$

   $\left \{ {{y>0} \atop {y∈Z}} \right.⇒\left \{ {{3-y<3} \atop {y∈Z}} \right.$

   `⇒(x-3);(3-y)∈{(1;-16);(2;-8);(4;-4);(8;-2);(16;-1)}`

   `⇒(x;y)∈{(4;19);(5;11);(7;7);(11;5);(19;4)}`.

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