\(S=\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+...+\dfrac{2}{98\cdot100}\)

\(S=\dfrac{2}{10}-\dfrac{2}{12}+\dfrac{2}{12}-\dfrac{2}{14}+...+\dfrac{2}{98}-\dfrac{2}{100}\)

\(S=\dfrac{2}{10}-\dfrac{2}{100}=\dfrac{9}{50}=0,18\)

Vậy \(S>\dfrac{1}{10}\)

\(S=\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+\dfrac{2}{14\cdot16}+...+\dfrac{2}{98\cdot100}\)

\(S=\dfrac{2}{10}-\dfrac{2}{12}+\dfrac{2}{12}-\dfrac{2}{14}+...+\dfrac{2}{98}-\dfrac{2}{100}\)

\(S=\dfrac{2}{10}-\dfrac{2}{100}\)

\(S=\dfrac{20}{100}-\dfrac{2}{100}\)

\(S=\dfrac{18}{100}=\dfrac{9}{50}=0,18\)

\(\dfrac{1}{10}=0,1\), mà \(0,1< 0,18\)

 \(\Rightarrow S>\dfrac{1}{10}\left(đpcm\right)\)

S=\(\dfrac{2}{10.12}\)+\(\dfrac{2}{12.14}\)+...+\(\dfrac{1}{98\cdot100}\)

S=\(\dfrac{1}{10}\)-\(\dfrac{1}{12}\)+\(\dfrac{1}{12}\)-\(\dfrac{1}{14}\)+...+\(\dfrac{1}{98}\)-\(\dfrac{1}{100}\)

S=\(\dfrac{1}{10}\)-\(\dfrac{1}{100}\)

S=\(\dfrac{9}{100}\)

Ta có \(\dfrac{1}{10}\)=\(\dfrac{10}{100}\)

mà 10>9

Suy ra \(\dfrac{9}{100}\)<\(\dfrac{10}{100}\)

hay \(\dfrac{9}{100}\)<\(\dfrac{1}{10}\) <=>S<\(\dfrac{1}{10}\) (đpcm)

(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 \(n\in\left\{3;1;4;0;6;-2\right\}\)

S=\(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+.....+\frac{2}{98.100}\)

S=\(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+........+\frac{1}{98}-\frac{1}{100}\)

S=\(\frac{1}{10}-\frac{1}{100}\)

S=\(\frac{9}{100}\)<\(\frac{1}{10}\)

tính S = cánh tính sai phân  

=1/2 - 1/4 + 1/4 - 1/6 + ... + 1/98 - 1/100

=1/2 - 1/100 = 49/100

1/2 - 1/4 +  1/4 - 1/6 + 1/6 - 1/8 + ... + 1/96 - 1/98 + 1/98 - 1/100

= 1/2 - 1/100 

= 49/100

1/2.4 + 1/4.6 + 1/6.8 + ... + 1/96.98 + 1/98.100

= 1/2 . ( 2/2.4 + 2/4.6 + 2/6.8 + ... + 2/96.98 + 2/98.100 )

= 1/2 . ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/96 - 1/98 + 1/98 - 1/100 )

= 1/2 . ( 1/2 - 1/100 )

= 1/2 . ( 50/100 - 1/100 )

= 49/200

bn lấy 1/2 nhân ra ngoài ròi tính như bình thường nha!

Đặt tổng trên là A ta có

\(2A=\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{48.52}\)

\(2A=\frac{12-10}{10.12}+\frac{14-12}{12.14}+\frac{16-14}{14.16}+...+\frac{50-48}{48.50}\)

\(2A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{48}-\frac{1}{50}=\frac{1}{10}-\frac{1}{50}=\frac{2}{25}\)

\(\Rightarrow A=\frac{2A}{2}=\frac{1}{25}\)
 

\(\frac{1}{10.12}+\frac{1}{12.14}+\frac{1}{14.16}+...+\frac{1}{48.50}\)

=\(\frac{1}{5.2.2.6}+\frac{1}{6.2.2.7}+\frac{1}{7.2.2.8}+...+\frac{1}{24.2.2.25}\)

=\(\frac{1}{2}.\left(\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{24.25}\right)\)

=\(\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\right)\)

=\(\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{25}\right)\)

=\(\frac{1}{2}.\frac{4}{25}\)

=\(\frac{2}{25}\)

mình không biết đúng hông có gì sai cho mình xin lỗi

nhóm từng cái vào r` tính ra

Đặt biểu thức cần tính là A, ta có:

A=\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}\right)\)

Làm tg tự với những cái khác là ok

Đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{98.100}\)

A=\(\left(\dfrac{1}{1.3}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+...+\dfrac{1}{98.100}\right)\)

A=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right)+\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

A=\(\dfrac{98}{99}-\dfrac{49}{100}\)

A=\(\dfrac{4949}{9900}\)

Mà \(\dfrac{3}{4}=\dfrac{7425}{9900}\)

Vậy A<\(\dfrac{3}{4}\)

Bạn hãy tính \(\dfrac{1}{1.3}+...+\dfrac{1}{98.100}\)= \(\dfrac{4949}{9900}\) sau đo chỉ cần chứng minh nó nhỏ hơn bằng cách quy đồng .

\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)

\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\)

\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)

\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)

\(\Rightarrow G=\dfrac{2}{3}.\dfrac{96}{505}\)

\(\Rightarrow G=\dfrac{64}{505}\)

giải hộ với

\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\\ G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{101}{505}-\dfrac{5}{505}\right)\\ G=\dfrac{2}{3}.\dfrac{96}{505}\\ G=\dfrac{64}{505}\)

`A=4/10.12+4/12.14+4/14.16+...+4/96.98`

`=2(2/10.12+2/12.14+2/14.16+...+2/96.98)`

`=2(1/10-1/12+1/12-1/14+1/14-1/16+...+1/96-1/98)`

`=2(1/10-1/98)`

`=2 . 22/245`

`=44/245`

`A=4/(10.12) + 4/(12.14) +  ... + 4/(96.98)`

`= 2 . (2/(10.12) + 2/(12.14) +  ... + 2/(96.98))`

`= 2 . (1/10 - 1/12 + 1/12 - 1/14 +....+ 1/96 - 1/98)`

`= 2. (1/10 - 1/98)`

`= 2 . 22/245`

`= 44/245`