Question

Tìm n thuộc Z để n² +13n - 13 chia hết cho n + 3

Answer 1

n²+13-13 chia hết cho n+3

=> n²-3²+3² chia het cho n+3

=> (n+3)(n-3)+9 chia het cho n+3

Vi (n+3)(n-3) chia het cho n+3 nen 9 chia het cho n+3

=> n+3 thuoc{+1;-1;+3;-3;+9;-9}

=> n thuoc {-2;-4;0;-6;6;-12}

Answer 2

n thuộc {-2;4;0;-6;6;-12}

Answer 3

Tìm n thuộc Z để n² +13n - 13 chia hết cho n + 3

Trả lời:

n² + 13 - 13 \(⋮\)n + 3

\(\Rightarrow\)n² - 3² + 3² \(⋮\)n + 3

\(\Rightarrow\)( n + 3 ) (  n - 3 ) + 9 \(⋮\)n + 3

Vì ( n + 3 ) (  n - 3 ) ​\(⋮\)chia hết cho n + 3 nên 9 \(⋮\)n + 3​

\(\Rightarrow n+3\in\left(+1;-1;+3;-3;+9;-9\right)\)

\(\Rightarrow n\in\left\{-2;-4;0;-6;6;-12\right\}\)

\(⋮\)

Answer 4

n² + 13 - 13 ⋮n + 3

⇒n² - 3² + 3² ⋮n + 3

⇒( n + 3 ) (  n - 3 ) + 9 ⋮n + 3

Vì ( n + 3 ) (  n - 3 ) ​⋮chia hết cho n + 3 nên 9 ⋮n + 3​

⇒n+3∈(+1;−1;+3;−3;+9;−9)

⇒n∈{−2;−4;0;−6;6;−12}

⋮

Answer 5

\(\left(n^2+13n-13\right)⋮\left(n+3\right)\)

\(\Rightarrow\left(n^2+3n+10n+30-43\right)⋮\left(n+3\right)\)

\(\Rightarrow\left[n\left(n+3\right)+10\left(n+3\right)+43\right]⋮\left(n+3\right)\)

\(\Rightarrow\left[\left(10+n\right)\left(n+3\right)+43\right]⋮\left(n+3\right)\)

Vì \(\Rightarrow\left[\left(10+n\right)\left(n+3\right)\right]⋮\left(n+3\right)\) nên \(43⋮\left(n+3\right)\)

\(\Rightarrow n+3\inƯ\left(43\right)=\left\{\pm1;\pm43\right\}\)

Lập bảng:

\(n+3\)	\(-1\)	\(1\)	\(-43\)	\(43\)
\(n\)	\(-4\)	\(-2\)	\(-46\)	\(40\)

Vậy \(x\in\left\{-4;-2;-46;40\right\}\)