Question

Cho phân số \(A=\frac{2n-1}{n+3}\). Tìm n\(\in\)Z để A có giá trị nguyên.

Answer 1

Ta có:     \(A=\frac{2n-1}{n+3}=2-\frac{7}{n+3}\)

Để  A  nguyên  thì   \(7\)\(⋮\)\(n+3\)

\(\Rightarrow\)\(n+3\)\(\inƯ\left(7\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\)\(n\)\(=\left\{-10;-4;-2;4\right\}\)

Answer 2

\(A=\frac{2n-1}{n+3}\) có giá trị nguyên

\(\Leftrightarrow2n-1⋮n+3\)

\(\Rightarrow\left(2n+6\right)-6-1⋮n+3\)

\(\Rightarrow2\left(n+3\right)-7⋮n+3\)

           có \(2\left(n+3\right)⋮n+3\)

\(\Rightarrow-7⋮n+3\)

\(\Rightarrow n+3\inƯ\left(-7\right)\)

        \(n\in Z\Rightarrow n+3\in Z\)

\(\Rightarrow n+3\in\left\{-1;-7;1;7\right\}\)

\(\Rightarrow n\in\left\{-4;-10;-2;4\right\}\)

Answer 3

\(A=\frac{2n-1}{n+3}\)

\(A=\frac{2\left(n+3\right)-7}{n+3}\)

\(A=2-\frac{7}{n+3}\)

để \(A\in Z\)thì \(\frac{7}{n+3}\in Z\)

\(\Leftrightarrow n+3\inƯ\left(7\right)\)

\(\Leftrightarrow n+3\in\left\{\pm1;\pm7\right\}\)

+ \(n+3=-1\Leftrightarrow n=-4\)

+ \(n+3=1\Leftrightarrow n=-2\)

+ \(n+3=7\Leftrightarrow n=4\)

+ \(n+3=-7\Leftrightarrow n=-10\)

vậy \(x\in\left\{\pm4;-2;-10\right\}\)

Answer 4

Để A có giá trị nguyên

\(\Rightarrow2n-1\) chia hết cho \(n+3\)

\(\Rightarrow n+3+n+3+7\) chia hết cho \(n+3\)

\(\Rightarrow\) 7 chia hết cho n + 3

\(\Rightarrow n+3\inƯ\left(7\right)\)

Mà \(Ư\left(7\right)=\left\{1;7;-1;-7\right\}\)

\(\Rightarrow n+3\in\left\{1;7;-1;-7\right\}\)

\(\Rightarrow n\in\left\{-2;4;-4;-10\right\}\)

Answer 5

Để A là số nguyên thì 2n - 1 ⋮ n + 3

<=> 2(n + 3) - 7 ⋮ n + 3

<=> 7 ⋮ n + 3 (vì 2(n + 3) ⋮ n + 3)

<=> n + 3 ∈ Ư(7) = {1; -1; 7; -7}

n + 3 = 1 => n = -2

n + 3 = -1 => n = -4

n + 3 = 7 => n = 4

n + 3 = -7 => n = -10

Vậy n ∈ {-2; -4; 4; -10}