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}

Ta có: \(\frac{n}{2n+1}=\frac{3n}{6n+3}\)

Vì 3n < 3n + 1 nên \(\frac{3n}{6n+3}<\frac{3n+1}{6n+3}\)

Vậy \(\frac{n}{2n+1}<\frac{3n+1}{6n+3}\)

Ta có:

n/2n + 1 = 3n/6n + 3

3n/6n + 3 < 3n + 1/6n + 3

=>n/2n + 1 <3n + 1/6n + 3

Thanks!

A = n/2n+1 = 3n / 6n+3 < 3n+1/6n+3 = B

=> A < B

\(\frac{n}{2n+1}\)=\(\frac{3.n}{3.\left(2n+1\right)}\)=\(\frac{3n}{6n+3}\)

Vì 6n+3=6n+3;3n<3n+1 nên \(\frac{n}{2n+1}\)<\(\frac{3n+1}{6n+3}\)

\(A=\dfrac{n}{2n+1}=\dfrac{n\left(6n+3\right)}{\left(2n+1\right)\left(6n+3\right)}\dfrac{6n^2+3n}{\left(2n+1\right)\left(6n+3\right)}\)

\(B=\dfrac{3n+1}{6n+3}=\dfrac{\left(3n+1\right)\left(2n+1\right)}{\left(6n+3\right)\left(2n+1\right)}=\dfrac{6n^2+5n+1}{\left(6n+3\right)\left(2n+1\right)}\)

Lại có :

\(6n^2+3n< 6n^2+5n+1\)

\(\Leftrightarrow A< B\)

$A=\frac{n}{2n+1}=\frac{n\left(6n+3\right)}{\left(2n+1\right)\left(6n+3\right)}\frac{6n^2+3n}{\left(2n+1\right)\left(6n+3\right)}$

$B=\frac{3n+1}{6n+3}=\frac{\left(3n+1\right)\left(2n+1\right)}{\left(6n+3\right)\left(2n+1\right)}=\frac{6n^2+5n+1}{(6}$