Question

a)Tìm nϵZ(3n+2) chia hết cho (n+1)

b)  (x-5).y=x+1(x,yϵZ)

 

HELP ME :

Answer 1

a: =>3n+3-1 chia hết cho n+1

=>\(n+1\in\left\{1;-1\right\}\)

=>\(n\in\left\{0;-2\right\}\)

b: =>xy-5y-x-1=0

=>x(y-1)-5y+5-6=0

=>(x-5)(y-1)=6

\(\Leftrightarrow\left(x-5;y-1\right)\in\left\{\left(1;6\right);\left(6;1\right);\left(-1;-6\right);\left(-6;-2\right);\left(2;3\right);\left(3;2\right);\left(-3;-2\right);\left(-2;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(6;7\right);\left(11;2\right);\left(4;-5\right);\left(-1;-1\right);\left(7;4\right);\left(8;3\right);\left(3;-2\right);\left(2;-1\right)\right\}\)