Tìm số tự nhiên n để 6n+3 chia hết 3n+6. 18/10/2021 Bởi Daisy Tìm số tự nhiên n để 6n+3 chia hết 3n+6.
`6n+3 \vdots 3n+6` `2(3n+6)-9 \vdots 3n+6` `⇒9 \vdots 3n+6` `⇒3n+6∈{±9;±3±1}` `⇒3(n+2)∈{±9;±3±1} ` `⇒n+2∈{±3;±1;±1/3} ` vì `n∈Z` `⇒n+2∈{±3;±1}` `⇒n∈{1;-1;-3;-5}` Bình luận
Đáp án: Giải thích các bước giải: Ta có: $\frac{6n+3}{3n+6}$=$\frac{6n+12-9}{3n+6}$=2-$\frac{9}{3n+6}$ để 6n+3:3n+6 ⇔9:3n+6 ⇒\(\left[ \begin{array}{l}3n+6=1\\3n+6=-1\end{array} \right.\) \(\left[ \begin{array}{l}3n+6=9\\3n+6=-9\end{array} \right.\) \(\left[ \begin{array}{l}3n+6=3\\3n+6=-3\end{array} \right.\) ⇒\(\left[ \begin{array}{l}3n=-5\\3n=-7\end{array} \right.\) \(\left[ \begin{array}{l}3n=3\\3n=-15\end{array} \right.\) \(\left[ \begin{array}{l}3n=-3\\3n=-9\end{array} \right.\) ⇒n=$\frac{-5}{3}$;$\frac{-7}{3}$;1;-5;-1;-3 Bình luận
`6n+3 \vdots 3n+6`
`2(3n+6)-9 \vdots 3n+6`
`⇒9 \vdots 3n+6`
`⇒3n+6∈{±9;±3±1}`
`⇒3(n+2)∈{±9;±3±1} `
`⇒n+2∈{±3;±1;±1/3} `
vì `n∈Z`
`⇒n+2∈{±3;±1}`
`⇒n∈{1;-1;-3;-5}`
Đáp án:
Giải thích các bước giải:
Ta có:
$\frac{6n+3}{3n+6}$=$\frac{6n+12-9}{3n+6}$=2-$\frac{9}{3n+6}$
để 6n+3:3n+6
⇔9:3n+6
⇒\(\left[ \begin{array}{l}3n+6=1\\3n+6=-1\end{array} \right.\)
\(\left[ \begin{array}{l}3n+6=9\\3n+6=-9\end{array} \right.\)
\(\left[ \begin{array}{l}3n+6=3\\3n+6=-3\end{array} \right.\)
⇒\(\left[ \begin{array}{l}3n=-5\\3n=-7\end{array} \right.\)
\(\left[ \begin{array}{l}3n=3\\3n=-15\end{array} \right.\)
\(\left[ \begin{array}{l}3n=-3\\3n=-9\end{array} \right.\)
⇒n=$\frac{-5}{3}$;$\frac{-7}{3}$;1;-5;-1;-3