Ta có: \(\frac{1}{1.2}=\frac{3}{1.2.3}\) ;\(\frac{1}{1.2+2.3}=\frac{3}{2.3.4}\); \(\frac{1}{2.3+3.4}=\frac{3}{3.4.5}\); ......;\(\frac{1}{1.2+2.3+3.4+...+n\left(n+1\right)}=\frac{3}{n\left(n+1\right)\left(n+2\right)}\)

=> \(S=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)}\)

=> \(\frac{2S}{3}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

Ta lại có: \(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\); \(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\); \(\frac{2}{3.4.5}=\frac{1}{3.4}-\frac{1}{4.5}\);....;\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)=> \(S=\frac{3}{4}-\frac{3}{2\left(n+1\right)\left(n+2\right)}< \frac{3}{4}\)

=> \(S< \frac{3}{4}\)

Mình nhầm 1 chỗ: \(\frac{1}{1.2+2.3+3.4}=\frac{3}{3.4.5}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{1999}-\frac{1}{2000}\)

\(=1-\frac{1}{2000}\)

\(=\frac{1999}{2000}\)

chịch

A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102

A=1-1/102=102/102-1/102=101/102

ý b thì chờ mình tí tìm cách lập luận đã nhé

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)

\(A=1-\frac{1}{102}\)

\(A=\frac{101}{102}\)

\(S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)

Vậy \(S< 1\left(đpcm\right)\)

1) \(\frac{1}{5}-\frac{1}{6}=\frac{6-5}{5.6}=\frac{1}{5.6}\)

\(\frac{1}{6}-\frac{1}{7}=\frac{7-6}{6.7}=\frac{1}{6.7}\)

2) Áp dụng bài trên, ta có:

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

= \(1-\frac{1}{6}=\frac{5}{6}\)

Câu 2: 

a: =>-11/12x=-1/6-3/4=-2/12-9/12=-11/12

=>x=1

b: =>x-42=57-x-50=7-x

=>2x=49

hay x=49/2

d: =>x+1=3 hoặc x+1=-3

=>x=2 hoặc x=-4

e: =>2x+3=5 hoặc 2x+3=-5

=>2x=2 hoặc 2x=-8

=>x=1 hoặc x=-4

Đặt Q = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{997.998}+\frac{1}{999.1000}\)

Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{997.999}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{997}-\frac{1}{999}\)

\(2A=1-\frac{1}{999}\)

\(2A=\frac{998}{999}\)

\(\Leftrightarrow A=\frac{499}{999}\)

Đặt B = \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{998.1000}\)

\(2B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{998}-\frac{1}{1000}\)

\(2B=\frac{1}{2}-\frac{1}{1000}\)

\(B=\frac{499}{1000}\)

Vậy Q = A + B = \(\frac{499}{999}+\frac{499}{1000}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{999.1000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{999}-\frac{1}{1000}\)

\(=1-\frac{1}{1000}=\frac{999}{1000}\)

Đặt 

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{999.1000}\)

\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)

\(\Leftrightarrow A=1-\frac{1}{1000}\)

\(\Leftrightarrow A=\frac{999}{1000}\)