S = \(\dfrac{1}{10.12}\) + \(\dfrac{1}{12.14}\) + .....+ \(\dfrac{1}{998.1000}\)

S = \(\dfrac{1}{2}\).( \(\dfrac{2}{10.12}\) + \(\dfrac{2}{12.14}\)+.....+ \(\dfrac{2}{998.1000}\))

S = \(\dfrac{1}{2}\).( \(\dfrac{1}{10}\)- \(\dfrac{1}{12}\)+ \(\dfrac{1}{12}\) - \(\dfrac{1}{14}\)+...+\(\dfrac{1}{998}\)- \(\dfrac{1}{1000}\))

S = \(\dfrac{1}{2}\). ( \(\dfrac{1}{10}\) - \(\dfrac{1}{1000}\))

S = \(\dfrac{1}{2}\).\(\dfrac{99}{1000}\)

S = \(\dfrac{99}{2000}\)

a) A = 3/10.12 + 3/12.14 + ... + 3/998.1000

2/3.A = 2/10.12 + 2/12.14 + ... + 2/998.1000

2/3.A = 1/10 - 1/12 + 1/12 - 1/14 + ... + 1/998 - 1/1000

2/3.A = 1/10 - 1/1000

2/3.A = 99/1000

A = 99/1000 : 2/3

A = 99/1000 . 3/2

A = 297/2000

b) B = 2/1.4 + 2/4.7 + 2/7.10 + ... + 2/22.25

3/2.B = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/22.25

3/2.B = 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/22 - 1/25

3/2.B = 1 - 1/25

3/2.B = 24/25

B = 24/25 : 3/2

B = 24/25 . 2/3

B = 16/25

Ủng hộ mk nha ^_-

a) Ta có: \(A=\frac{3}{10.12}+\frac{3}{12.14}+....+\frac{3}{998.1000}.\)

 \(\Rightarrow\frac{2}{3}A=\frac{1}{10.12}+\frac{1}{12.14}+...+\frac{1}{998.1000}\)

 \(\Rightarrow\frac{2}{3}A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{998}-\frac{1}{1000}\)

 \(\Rightarrow\frac{2}{3}A=\frac{1}{10}-\frac{1}{1000}=\frac{99}{1000}\)

 \(\Rightarrow A=\frac{99}{1000}:\frac{2}{3}=\frac{297}{2000}\)

Ta có: \(B=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{22.25}\)

\(\Rightarrow\frac{3}{2}B=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}\)

\(\Rightarrow\frac{3}{2}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{22}-\frac{1}{25}\)

\(\Rightarrow\frac{3}{2}B=1-\frac{1}{25}=\frac{24}{25}\)

\(\Rightarrow B=\frac{24}{25}:\frac{3}{2}=\frac{16}{25}\)

Đặt \(A=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{48\cdot50}\)

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

\(A=\frac{1}{10}-\frac{1}{50}=\frac{5}{50}-\frac{1}{50}=\frac{4}{50}=\frac{2}{25}\)

Vậy \(A=\frac{2}{25}\)

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

2/10.12+2/12.14+2/14.16+...+2/48.50

=1/10-1/12+1/12-1/14+1/14-1/16+...+1/48-1/50

=( 1/10-1/50 )+( 1/12-1/12 )+( 1/14-1/14 )+...+( 1/48-1/48 )

=( 5/50-1/50 )+0+0+...+0

=4/50

=2/25

CHÚC BẠN THÂN YÊU HỌC TỐT ^:^

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  

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

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

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

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

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)