Bài 2: 

a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)

\(=\dfrac{4+6-3}{n-1}\)

\(=\dfrac{7}{n-1}\)

Để A là số tự nhiên thì \(7⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(7\right)\)

\(\Leftrightarrow n-1\in\left\{1;7\right\}\)

hay \(n\in\left\{2;8\right\}\)

Vậy: \(n\in\left\{2;8\right\}\)

ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2                                                   Để B là STN thì 4n+10⋮n+2                          4n+8+2⋮n+2                                  4n+8⋮n+2                                                      ⇒2⋮n+2                                     n+2∈Ư(2)                                                        Ư(2)={1;2}                                  Vậy n=0                                                                                  

Ta có: =4+6−3n−1=4+6−3�−1

b: =>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{200}{101}\)

=>\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{100}{101}\)

=>1-1/2+1/2-1/3+...+1/n-1/n+1=100/101

=>1-1/(n+1)=100/101

=>1/(n+1)=1/101

=>n+1=101

=>n=100

a) \(\dfrac{n+2}{3}\) là số tự nhiên khi

\(n+2⋮3\)

\(\Rightarrow n+2\in\left\{1;3\right\}\)

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

b)  \(\dfrac{7}{n-1}\) là số tự nhiên khi

\(7⋮n-1\)

\(\Rightarrow7n-7\left(n-1\right)⋮n-1\)

\(\Rightarrow7n-7n+7⋮n-1\)

\(\Rightarrow7⋮n-1\)

\(\Rightarrow n-1\in\left\{1;7\right\}\Rightarrow\Rightarrow n\in\left\{2;8\right\}\left(n\in Z\right)\)

c) \(\dfrac{n+1}{n-1}\) là sô tự nhiên khi

\(n+1⋮n-1\)

\(\Rightarrow n+1-\left(n-1\right)⋮n-1\)

\(\Rightarrow n+1-n+1⋮n-1\)

\(\Rightarrow2⋮n-1\)

\(\Rightarrow n-1\in\left\{1;2\right\}\Rightarrow n\in\left\{2;3\right\}\left(n\in Z\right)\)

Ta có \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};...;\dfrac{1}{n^2}=\dfrac{1}{n.n}< \dfrac{1}{\left(n-1\right)n}\)

Do đó \(a< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}=1+\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+...+\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)

\(=1+1-\dfrac{1}{n}=1-\dfrac{1}{n}< 2\) . Suy ra \(1< a< 2\)

Vậy \(a\) khôg phải số tự nhiên

Ta có: `1 < 1 + 1/2^2 + ... + 1/n^2`

`1/(2.2) < 1/(1.2)`

`1/(3.3) < 1/(2.3)`

`...`

`1/(n^2) < 1/(n-1(n))`

`=> 1/2^2 + ... + 1/n^2 < 1/(1.2) + ... + 1/(n-1(n)) = 1/1 - 1/n < 1`.

`=> a < 1 + 1 = 2`.

`=> 1 < a < 2`.

`=>` Đây không là số tự nhiên.

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}\\ A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\left(\dfrac{1}{n}>0\right)\)

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

Để A là số tự nhiên => \(n+1\inƯ\left(2\right)=\left\{-2,-1,1,2\right\}\)

Vậy \(n\in\left\{0;2;3\right\}\)

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

2n-4+4⋮n-2

2n-4⋮n-2⇒4⋮n-2

n-2∈Ư(4)⇒Ư(4)={1;-1;2;-2;4;-4}

n∈{3;1;4;0;6;-2}

(2) \(\dfrac{3}{10.12}+\dfrac{3}{12.14}+...+\dfrac{3}{48.50}\)

=\(\dfrac{3}{2}.\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+...+\dfrac{2}{48.50}\right)\)

=\(\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

=\(\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{50}\right)\)

=\(\dfrac{3}{2}.\dfrac{2}{25}\)

=\(\dfrac{3}{25}\)

Giải:

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

\(2n⋮n-2\) 

\(\Rightarrow2n-4+4⋮n-2\) 

\(\Rightarrow4⋮n-2\) 

\(\Rightarrow n-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\) 

Vậy \(n\in\left\{0;1;3;4;6\right\}\)

(2) \(\dfrac{3}{10.12}+\dfrac{3}{12.14}+\dfrac{3}{14.16}+...+\dfrac{3}{48.50}\) 

\(=\dfrac{3}{2}.\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+\dfrac{2}{14.16}+...+\dfrac{2}{48.50}\right)\) 

\(=\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\) 

\(=\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{50}\right)\) 

\(=\dfrac{3}{2}.\dfrac{2}{25}\) 

\(=\dfrac{3}{25}\) 

Chúc bạn học tốt!

(1) Để biểu thức \(\dfrac{2n}{n-2}\) nguyên thì \(2n⋮n-2\)

\(\Leftrightarrow4⋮n-2\)

\(\Leftrightarrow n-2\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{3;1;4;0;6;-2\right\}\)