Question

Tìm n thuộc Z biết 

a,n+4 chia hết n+3

b,n-5 chia hết n+6

c,2n-7 chia hết n+4

Answer 1

\(a,n+4⋮n+3\)

\(\Leftrightarrow\left(n+3\right)+1⋮n+3\)

Vì \(n+3⋮n+3\Rightarrow1⋮n+3\)

\(\Leftrightarrow n+3\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(\Leftrightarrow n\in\left\{-4;-2\right\}\)

Vậy \(n\in\left\{-4;-2\right\}.\)

b,\(n-5⋮n+6\)

\(\Leftrightarrow\left(n+6\right)-11⋮n+6\)

Vì \(n+6⋮n+6\Rightarrow-11⋮n+6\)

\(\Rightarrow n+6\inƯ\left(-11\right)=\left\{-11;-1;1;11\right\}\)

\(\Rightarrow n\in\left\{-17;-7;-5;5\right\}\)

Vậy\(n\in\left\{-17;-7;-5;5\right\}.\)

c,\(2n-7⋮n+4\)

\(\Leftrightarrow\left(n+4\right)+\left(n+4\right)-15⋮n+4\)

Vì \(n+4⋮n+4\Rightarrow-15⋮n+4\)

\(\Rightarrow n+4\inƯ\left(-15\right)=\left\{-15;-5;-3;-1;1;3;5;15\right\}\)

\(\Rightarrow n\in\left\{-19;-9;-7;-5;-3;-1;1;11\right\}\)

Vậy\(n\in\left\{-19;-9;-7;-5;-3;-1;1;11\right\}.\)

Answer 2

a, n+4 chia hết cho n+3

=>n+3+1 chia hết cho n+3

=>1 chia hết cho n+3

=>n+3 E Ư(1)={1;-1}

=>n E {-2;-4}

b, n-5 chia hết cho n+6

=>n+6-11 chia hết cho n+6

=>11 chia hết cho n+6

=>n+6 E Ư(11)={1;-1;11;-11}

=>n  E {-5;-7;5;-17}

c,2n-7 chia hết cho n+4

=>2n+8-15 chia hết cho n+4

=>2(n+4)-15 chia hết cho n+4

=>15 chia hết cho n+4

=>n+4 E Ư(15)={1;-1;3;-3;5;-5;15;-15}

=>n E {-3;-5;-1;-7;1;-9;11;-19}

Answer 3

a. \(n+4⋮n+3\)

Mà \(n+3⋮n+3\)

\(\Leftrightarrow1⋮n+3\)

\(\Leftrightarrow n+3\inƯ\left(1\right)\)

\(\Leftrightarrow\orbr{\begin{cases}n+3=1\\n+3=-1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}n=-2\\n=-4\end{cases}}\)

Vậy ..........

b/ \(n-5⋮n+6\)

Mà \(n+6⋮n+6\)

\(\Leftrightarrow11⋮n+6\)

\(\Leftrightarrow x+6\inƯ\left(11\right)\)

Suy ra :

+) x + 6 = 1 => x = -5

+) x + 6 = 11 => x = 5