\(a)\,\,A=\dfrac{13}{21} \Leftrightarrow \dfrac{2n+3}{4n+1}=\dfrac{13}{21} \\ \Leftrightarrow 21(2n+3)=13(4n+1)\\\Leftrightarrow 42n+63=52n+13\\\Leftrightarrow 42n-52n=13-63 \\\Leftrightarrow -10n=-50\\\Leftrightarrow n=(-50):(-10)\\\Leftrightarrow n=5\)

a, \(A=\dfrac{5n-4-4n+5}{n-3}=\dfrac{n+1}{n-3}=\dfrac{n-3+4}{n-3}=1+\dfrac{4}{n-3}\Rightarrow n-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

a.\(A=\dfrac{2n+1}{n-3}+\dfrac{3n-5}{n-3}-\dfrac{4n-5}{n-3}\)

\(A=\dfrac{2n+1+3n-5-4n+5}{n-3}\)

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

\(A=\dfrac{n-3}{n-3}+\dfrac{4}{n-3}\)

\(A=1+\dfrac{4}{n-3}\)

Để A nguyên thì \(\dfrac{4}{n-3}\in Z\) hay \(n-3\in U\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

n-3=1 --> n=4

n-3=-1 --> n=2

n-3=2 --> n=5

n-3=-2 --> n=1

n-3=4 --> n=7

n-3=-4 --> n=-1

Vậy \(n=\left\{4;2;5;7;1;-1\right\}\) thì A nhận giá trị nguyên

b.hemm bt lèm:vv

cc

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

A \(\in\) Z ⇔ 3n + 1 ⋮ 2n + 3

             6n + 2 ⋮ 2n + 3

         6n + 9 - 7 ⋮ 2n + 3

    3.(2n + 3) - 7 ⋮ 2n + 3

                      7 ⋮ 2n + 3 ⇒ 2n + 3 \(\in\) Ư(7) = { -7; -1; 1; 7}

Lập bảng ta có: 

Vậy các số nguyên n thỏa mãn đề bài là:

n \(\in\) { -5; -2; -1; 2}

\(A=\dfrac{3n+1}{2n+3}\inℤ\) \(\left(n\ne-\dfrac{3}{2}\right)\)

\(\Rightarrow3n+1⋮2n+3\)

\(\Rightarrow2\left(3n+1\right)-3\left(2n+3\right)⋮2n+3\)

\(\Rightarrow6n+2-6n-9⋮2n+3\)

\(\Rightarrow-7⋮2n+3\)

\(\Rightarrow2n+3\in\left\{-1;1;-7;7\right\}\)

\(\Rightarrow n\in\left\{-2;-1;-5;2\right\}\)

Để A là số nguyên thì 2n+2-5 chia hết cho n+1

=>\(n+1\in\left\{1;-5;-1;5\right\}\)

=>\(n\in\left\{0;-6;-2;4\right\}\)

ta có A=\(\frac{n+1}{n-3}\)

để A nguyên thì \(n+1⋮n-3\Rightarrow n-3+4⋮̸n-3\)

vì \(n-3⋮n-3\Rightarrow4⋮n-3\Rightarrow\left(n-3\right)\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

vậy \(n\in\left\{2;1;-1;4;5;7\right\}\)


 

a) \(a\left(b+1\right)=3\left(a;b\inℤ\right)\)

\(\Rightarrow a;\left(b+1\right)\in U\left(3\right)=\left\{-1;1;-3;3\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(-1;-4\right);\left(1;2\right);\left(-3;-2\right);\left(3;0\right)\right\}\)

b) \(2n+7⋮n+1\left(n\inℤ\right)\)

\(\Rightarrow2n+7-2\left(n+1\right)⋮n+1\)

\(\Rightarrow2n+7-2n-2⋮n+1\)

\(\Rightarrow5⋮n+1\)

\(\Rightarrow n+1\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

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

c) \(xy+x-y=6\left(x;y\inℤ\right)\)

\(\Rightarrow x\left(y+1\right)-y-1+1=6\)

\(\Rightarrow x\left(y+1\right)-\left(y+1\right)=5\)

\(\Rightarrow\left(x-1\right)\left(y+1\right)=5\)

\(\Rightarrow\left(x-1\right);\left(y+1\right)\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-0;-6\right);\left(2;4\right);\left(-4;-2\right);\left(6;0\right)\right\}\)

Viết  lời giải ra giúp mình nhé !