A = 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100

A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +.....+ 1/99- 1/100

A= 1 - 1/100

A= 99/100

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ghi xong hết rồi

mạng nó rớt, ấn gửi trả lời mà không biết

tong teo

a)A = 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100

A = 1 -1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

Rút gọn ta được :

A= 1 - 1/100

A= 99/100

b) B = 1/1.3+1/3.5+1/5.7+....+1/ 99 .101

   B x 2 ta có : 1- 1/3 + 1/3 - 1/5+ 1/5-1/7+...+1/99-1/101

   B x2 rút gọn ta được: 1 - 1/ 101 

                         B x 2=  100 / 101

                        B =  100/ 101 : 2 = 50 / 101

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

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

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

\(=\frac{99}{100}\)

Sửa lại đề bài nha bạn:

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

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

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

Chúc em học tốt nhé!

Xin phép sửa đề 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/99.100

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

A=1-1/100                            A=99/100                                                                                    B= (1/5.6+1/6/7+...+1/101.102).3                         B=(1/5-1/6+1/6-1/7+...+1/101-1/102).3        B=(1/5-1/102).3                                                 B=97/170                                                            

1) Tính

a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

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

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

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

\(=\frac{99}{100}\)

b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2.\left(1-\frac{1}{99}\right)\)

\(=2.\frac{98}{99}\)

\(=\frac{196}{99}=1\frac{97}{99}\)

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

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

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

\(=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)

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

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

\(=\frac{98}{99}\)

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

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

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

=>\(\frac{99}{100}\)

B=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\)

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

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

=>\(\frac{98}{99}\)

B = 1.2+2.3+3.4+...+99.100

B=1.100

B=100

C=1.3+2.4+3.5+4.6+...+9.11

C=1.(2+1)+2.(3+1)+3.(4+1)+4.(5+1)+...+9.(10+1)

C=1.2+1+2.3+1+3.4+1+4.5+1+...+9.10+1

C=(1.2+2.3+3.3+4.5+...+9.10)+(1+1+1+1+..+1)

C=1.10+10

C=10+10

C=20

a) B = 1.2+2.3+3.4+..+99.100

=>3B=1.2.3+2.3.3+3.4.3+...+99.100.3

3B = 1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5-2.3.4+...+99.100.101-98.99.100

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

3B = 99.100.101

\(B=\frac{99.100.101}{3}=333300\)

b) C = 1.3+2.4+3.5+4.6+...+9.11

C = (2-1).(2+1)+(3-1).(3+1) + (4-1).(4+1)+(5-1).(5+1)+...+(10-1).(10+1)

C = 2² - 1 + 3² - 1 + 4² - 1 + 5² - 1 +...+10² - 1

C = (2²+3²+4²+5²+...+10²) -(1+1+...+1) 

...

A = 1.2. + 2.3 + 3.4 + ... + 99.100

3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)

3A = 1.2.3 + 2.3.4 - 2.3.1 + ... + 99.100.101 - 99.100.98

3A = 99.100.101

3A = 999900

A = 333300

lấy nick khác hả không qua được mắt tui đâu đồ bất công