\(\frac{4n-5}{2n-1}=\frac{2\left(2n-1\right)-3}{2n-1}=2-\frac{3}{2n-1}\)

Vậy để 4n-5 chia hết cho 2n-1 thì \(2n-1\inƯ\left(3\right)\)

Mà Ư(3)={-1;1;3;-3}

+)2n-1=1 <=> n=1

+)2n-1=-1 <=> n=0

+)2n-1=3 <=> n=2

+)2n-1=-3 <=> n=-1

Vậy n={-1;0;1;2}

\(\frac{4n-5}{2n-1}=\frac{2\left(2n-1\right)}{2n-1}=\frac{2\left(2n-1\right)-3}{2n-1}=\frac{2\left(2n-1\right)}{2n-1}-\frac{3}{2n-1}=2-\frac{3}{2n-1}\in Z\)

\(\Rightarrow3⋮2n-1\)

\(\Rightarrow2n-1\inƯ\left(3\right)=\left\{1;3\right\}\left(n\in N\right)\)

\(\Rightarrow2n\in\left\{2;4\right\}\)

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

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

a: ĐKXĐ: n<>1

Để \(\frac{2n-1}{n-1}\) là số nguyên thì 2n-1⋮n-1

=>2n-2+1⋮n-1

=>1⋮n-1

=>n-1∈{1;-1}

=>n∈{2;0}

b: ĐKXĐ: n<>-1

Để \(\frac{3n+5}{n+1}\) là số nguyên thì 3n+5⋮n+1

=>3n+3+2⋮n+1

=>2⋮n+1

=>n+1∈{1;-1;2;-2}

=>n∈{0;-2;1;-3}

c: ĐKXĐ: n<>-3

Để \(\frac{4n-2}{n+3}\) là số nguyên thì 4n-2⋮n+3

=>4n+12-14⋮n+3

=>-14⋮n+3

=>n+3∈{1;-1;2;-2;7;-7;14;-14}

=>n∈{-2;-4;-1;-5;4;-10;11;-17}

d: ĐKXĐ: n<>-4/3

Để \(\frac{6n-4}{3n+4}\) là số nguyên thì 6n-4⋮3n+4

=>6n+8-12⋮3n+4

=>-12⋮3n+4

=>3n+4∈{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}

=>3n∈{-3;-5;-2;-6;-1;-7;0;-8;2;-10;8;-16}

=>n∈{\(-1;-\frac53;-\frac23;-2;-\frac13;-\frac73;0;-\frac83;\frac23;-\frac{10}{3};\frac83;-\frac{16}{3}\) }

mà n là số nguyên

nên n∈{-1;-2;0}

e: ĐKXĐ: n<>1/2

Để \(\frac{n+3}{2n-1}\) là số nguyên thì n+3⋮2n-1

=>2n+6⋮2n-1

=>2n-1+7⋮2n-1

=>7⋮2n-1

=>2n-1∈{1;-1;7;-7}

=>2n∈{2;0;8;-6}

=>n∈{1;0;4;-3}

f: \(\frac{6n-4}{3n-2}=\frac{2\left(3n-2\right)}{3n-2}=2\) là số nguyên với mọi n nguyên

g: ĐKXĐ: n<>1/3

Để \(\frac{2n+3}{3n-1}\) là số nguyên thì 2n+3⋮3n-1

=>6n+9⋮3n-1

=>6n-2+11⋮3n-1

=>11⋮3n-1

=>3n-1∈{1;-1;11;-11}

=>3n∈{2;0;12;-10}

=>n∈{2/3;0;4;-10/3}

mà n nguyên

nên n∈{0;4}

a: \(\Leftrightarrow2n-1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{1;0;2;-1\right\}\)

c: \(\Leftrightarrow n+1\in\left\{1;-1\right\}\)

hay \(n\in\left\{0;-2\right\}\)

<=>4n-5=4n-2+7

<=>2.(2n-1)+7

vì 2.(2n-1) chia hết cho 2n-1

Nên 2n-1 thuộc Ư(7)={1;7;-1;-7}

do đó 2n-1=1=>n=1

2n-1=7=>n=8

2n-1=-1=>n=0

2n-1=-7=>n=-3

Vậy n ={1;8;0;-3}

\(\frac{4n-5}{2n-1}=\frac{4n-2}{2n-1}-\frac{3}{2n-1}\)\(=\frac{2\left(2n-1\right)}{2n-1}-\frac{3}{2n-1}\)\(=2-\frac{3}{2n-1}\)

=> \(\frac{3}{2n-1}\in Z=>\)\(3⋮\left(2n-1\right)=>2n-1\inƯ\left(3\right)\)

=> \(2n-1\in\left\{-3;-1;1;3\right\}\)

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

=> n thuộc { -1;0;1;2}

a) n = 0 hoặc n= 2

n = -3 hoặc n=-1