A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

A = \(1-\frac{1}{100}=\frac{99}{100}\)

Bài 1

a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

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

\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{33}{99}-\frac{1}{99}\)

\(=\frac{32}{99}\)

c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{10}{20}-\frac{1}{20}\)

\(=\frac{9}{20}\)

Tk mình nha!!

Câu 2:

\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)

\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)

\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)

\(=\frac{3\cdot100}{2}\)

\(=\frac{300}{2}=150\)

BÀI 1:

\(N=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{17.20}\)

\(N=2.\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{17.20}\right)\)

\(N=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(N=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(N=2.\frac{9}{20}\)

\(N=\frac{9}{10}\)

BÀI 2:

\(C=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3B=1.2\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(3B=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+...+98.99.100\right)\)

\(3B=99.100.101\)

\(3B=999900\)

\(\Rightarrow B=999900:3\)

\(B=333300\)

CHÚC BN HỌC TỐT!!!!

Thanks CÔNG CHÚA ÔRI vì đã giúp mk

A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

   = \(1-\frac{1}{2017}\)

   = \(\frac{2016}{2017}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(A=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{2016}+\frac{1}{2016}\right)-\frac{1}{2017}\)

\(A=1+0+0+...+0-\frac{1}{2017}\)

\(A=1-\frac{1}{2017}\)

\(A=\frac{2017}{2017}-\frac{1}{2017}\)

\(A=\frac{2016}{2017}\)

Vậy:  \(A=\frac{2016}{2017}\)

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 

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

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

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

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

bạn viết sai đề rồi

Đặt biểu thức là A.

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

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

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