Toán Lớp 8: Tìm n để `103n^2 +121n +70` chia hết cho `n-1`

Question

Toán Lớp 8:  Tìm n để 103n^2 +121n +70 chia hết cho n-1

quynhmaithu · Answer

Giải đáp:   $n  in  {&minus;293,&minus;146,&minus;97,&minus;48,&minus;41,&minus;20,&minus;13,&minus;6,&minus;5,&minus;2,&minus;1,0,2,3,4,7,8,15,22,43,50,99,148,295 }.$   Lời giải và giải thích chi tiết:   $(103n^2+121n+70)    vdots    (n-1)    Rightarrow  dfrac{103n^2+121n+70}{n-1}  in  mathbb{Z}    Leftrightarrow  dfrac{103n^2-103n+224n-224+294}{n-1}  in  mathbb{Z}    Leftrightarrow  dfrac{103n(n-1)+224(n-1)+294}{n-1}  in  mathbb{Z}    Leftrightarrow 103n+224 +  dfrac{294}{n-1}  in  mathbb{Z}    Leftrightarrow   dfrac{294}{n-1}  in  mathbb{Z}( text{Do } n  in  mathbb{Z})   n  in  mathbb{Z},  dfrac{294}{n-1}   Rightarrow (n-1)  in Ư(294)    Leftrightarrow (n-1)  in  {&minus;294,&minus;147,&minus;98,&minus;49,&minus;42,&minus;21,&minus;14,&minus;7,&minus;6,&minus;3,&minus;2,&minus;1,1,2,3,6,7,14,21,42,49,98,147,294 }    Leftrightarrow n  in  {&minus;293,&minus;146,&minus;97,&minus;48,&minus;41,&minus;20,&minus;13,&minus;6,&minus;5,&minus;2,&minus;1,0,2,3,4,7,8,15,22,43,50,99,148,295 }.$