Bài 1 :

Đặt A=1.2+2.3+3.4+4.5+.........+99.100

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

3A=1.2.3+2.3.4-1.2.3+........+99.100.101-98.99.100

3A=99.100.101

A=33.100.101

A=333300

Bài 2 :

1:20 + 1:44 + 1:77 + 1:119 + 1:170 = \(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}=\frac{1}{10}=0,1\)

           1)1.2+2.3+3.4+4.5+...+99.100

          đặt 3D=1.2+2.3+3.4+...+99.100

                   =1.2.3+2.8.3+...+3.4.3+4.5.3+...+99.100.3

                   =1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

                  =1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5

                  =99.100.101

                 =999900

              D=999900:3=333300 

nếu đúng nhớ cảm ơn nhak. mình ko bít làm bài 2

\(\frac{1}{20}+\left(\frac{1}{44}+\frac{1}{77}\right)+\left(\frac{1}{119}+\frac{1}{170}\right)=\frac{1}{20}+\left(\frac{1}{11}.\frac{1}{4}+\frac{1}{11}.\frac{1}{7}\right)+\left(\frac{1}{17}.\frac{1}{7}+\frac{1}{17}.\frac{1}{10}\right)\)

= \(\frac{1}{20}+\frac{1}{11}.\left(\frac{1}{4}+\frac{1}{7}\right)+\frac{1}{17}.\left(\frac{1}{7}+\frac{1}{10}\right)=\frac{1}{20}+\frac{1}{11}.\frac{11}{28}+\frac{1}{17}.\frac{17}{70}=\frac{1}{20}+\frac{1}{28}+\frac{1}{70}\)

= \(\frac{1}{20}+\frac{1}{14}.\left(\frac{1}{2}+\frac{1}{5}\right)=\frac{1}{20}+\frac{1}{14}.\frac{7}{10}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=0,1\)

Lời giải :

Đặt S=1.2+2.3+3.4+4.5+…+99.100+100.101

3S=1.2.3+2.3.3+3.4.3+4.5.3+…+99.100.3+100.101.3

=1.2(3−0)+2.3(4−1)+3.4(5−2)+4.5(6−3)+…+99.100(101−98)+100.101(102−99)

=0.1.2-1.2.3+1.2.3-2.3.4+...+99.100.101-100.101.102

=100.101.102

S=100.101.34=343400

1.Tính 

a) Ta có: 

  A=(1-1/2²).(1-1/3²)...(1-1/100²)

=>A=3/2².8/3².....9999/100²

=>A=(1.3/2.2).(2.4/3.3).....(99.101/100.100)

=>A=(1.2.3.....99/2.3.4.....100).(3.4.5.....101/2.3.4.....100)

=>A=1/100.101/2

=>A=101/200

b) Ta có: 

  B=-1/1.2-1/2.3-1/3.4-...-1/100.101

=>B=-(1/1.2+1/2.3+1/3.4+...+1/100.101)

=>B=-(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)

=>B=-(1-1/101)

=>B=-100/101

 c) Ta có:

 C=1.2+2.3+3.4+...+100.101

       =>3C=1.2.3+2.3.3+3.4.3+...+100.101.3

       =>3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)

       =>3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102

       =>3C=100.101.102

       =>3C=1030200

       =>C=343400

Chúc bạn hok tốt nhé >:)!!!!!

cau hỏi tương tự ko có mà!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)

3C=2014.2015.2016

C=2014.2015.2016:3

bạn lên mạng tra từng câu 1 sẽ có

ukm cảm ơn bạn nhìu

Ta thấy:\(\frac{1}{1.2}=1-\frac{1}{2},\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3},...,\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)

=>\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

=>\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

=>\(A=1-\frac{1}{50}\)

=>\(A=\frac{49}{50}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow A=1-\frac{1}{50}\)

\(\Rightarrow A=\frac{49}{50}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\frac{1}{1}-\frac{1}{50}\)

\(A=\frac{50}{50}-\frac{1}{50}\)

\(A=\frac{49}{50}\)

v

Answer:

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

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

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

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

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

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(B=1.2+2.3+3.4+...+49.50\)

\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(=49.50.51\)

\(B=\frac{49.50.51}{3}=49.50.17\)

\(50^2.A-\frac{B}{17}=49.50-49.50=0\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\\ =1-\dfrac{1}{2014}\\ =\dfrac{2013}{2014}\)

A = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}...-\dfrac{1}{97.98}\)

A= 1-\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{97.98}\right)\)

A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)

A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{98}\right)\)

A=1-  1 + \(\dfrac{1}{98}\)

A= \(\dfrac{1}{98}\)

Lời giải:

$1-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{97.98}$

$1-A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{98-97}{97.98}$

$1-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}$

$=1-\frac{1}{98}$

$\Rightarrow A=\frac{1}{98}$