Toán Lớp 6: tìm các số tự nhiên n sao cho 3n+13 chia hết cho n+1

Question

Toán Lớp 6:  tìm các số tự nhiên n sao cho 3n+13 chia hết cho n+1

quynhmaithu · Answer

#andy     [ begin{array}{l} 3n + 13 { mathinner{ mkern2mu raise1pt hbox{.} mkern2mu   raise4pt hbox{.} mkern2mu raise7pt hbox{.} mkern1mu}} n + 1     Rightarrow 3 left( {n + 1}  right) + 10 { mathinner{ mkern2mu raise1pt hbox{.} mkern2mu   raise4pt hbox{.} mkern2mu raise7pt hbox{.} mkern1mu}} n + 1     Rightarrow 10 { mathinner{ mkern2mu raise1pt hbox{.} mkern2mu   raise4pt hbox{.} mkern2mu raise7pt hbox{.} mkern1mu}} n + 1     Rightarrow n + 1  in U { 10 }      Rightarrow U { 10 }  =  { 1; ,2;5:10 }      Rightarrow n + 1 =  { 1;2;5;10 }      Rightarrow n =  { 1 - 1;2 - 1;5 - 1;10 - 1 }      Rightarrow n =  { 0;1;4;9 }   end{array} ]

tonhitruc · Answer

- Ta c&oacute;: 3n + 13 ⋮ n + 1     => 2n + n + 1 + 12 ⋮ n + 1     2n + 12 ⋮ n + 1    1n + n + 1 + 11    ⋮   n + 1     n + 11 ⋮ n + 1     n + 1 + 10 ⋮ n + 1     10 ⋮ n + 1     n + 1 &isin; Ư(10) = { 1; 2; 5; 10 }     n &isin; { 1-1; 2-1; 5-1; 10-1 }     - Vậy: n &isin; { 0; 1; 4; 9 }     @ Rin