Có : (n + 5) vdots (n -2) => (n + 5) – (n-2) vdots (n-2) => n + 5 – n + 2 vdots (n-2) => 7 vdots ( n-2) => (n-2) in Ư(7) = {+-1;+-7} Lập bảng : \begin{array}{||c|c|}\hline{n-2} &\text{1}&\text{-1}&\text{7}&\text{-7}\\\hline \text{n}&\text{3<TM>}&\text{1<TM>}&\text{9<TM>}&\text{-5<TM>}\\\hline\end{array} => x in { 3 ; 1 ; 9 ; -5 } @121I Trả lời
n+5 \vdots n-2 => n-2+7 \vdots n-2 => 7 \vdots n-2 => n-2 in Ư(7) Mà Ư(7) = {+-1 ; +-7} Nên ta có : \begin{array}{|c|c|c|}\hline \text{n-2}&\text{1}&\text{-1}&\text{7}&\text{-7}\\\hline \text{n}&\text{3}&\text{1}&\text{9}&\text{-5}\\\hline\end{array} Trả lời
2 bình luận về “n+5 cho hết n-2 ( Lập Bảng )”