A = 1 + 1/110 + 1 + 1/90 + ... + 1  + 1 /2 

A = 10 + 1/1.2+ 1 /2.3 + ... + 1/9.10 + 1/10.11

A = 10 + 1/1 - 1/2 + 1 /2 - 1/3 + ... + 1/9 - 1/10 + 1/10 - 1/11

A = 10 + 1/1 - 1/11

A = 10 + 10/11

A = 120/11

A = \(\frac{111}{110}+\frac{91}{90}+\frac{73}{72}+...+\frac{13}{12}+\frac{7}{6}+\frac{3}{2}\)

A = \(\left(\frac{1}{2}+1\right)+\left(\frac{1}{6}+1\right)+\left(\frac{1}{12}+1\right)+....+\left(\frac{1}{110}+1\right)\)

A = (1 + 1 + 1 +...+ 1) + \(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)

A = 10 + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

A = \(10+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

A = \(10+\left(1-\frac{1}{11}\right)\)

A = \(10+\frac{10}{11}\)

A = \(\frac{120}{11}\)

hello bạn 

Gọi tổng dãy số hạng trên là A

A = 1 + \(\frac{1}{2}\)+ 1 + \(\frac{1}{6}\)+ 1 + \(\frac{1}{12}\)+ ... + 1 + \(\frac{1}{90}\)+ 1 + \(\frac{1}{110}\)

Mà từ \(\frac{1}{2}\)đén \(\frac{1}{110}\) có 10 số

A = 1 x 10 + \(\frac{1}{2}\)+( \(\frac{1}{2}\)- \(\frac{1}{3}\)) + ( \(\frac{1}{3}\)-\(\frac{1}{4}\)) + (\(\frac{1}{4}\)-\(\frac{1}{5}\)) + ... + \(\frac{1}{11}\) 

A = 10 + \(\frac{1}{2}\)+ \(\frac{1}{2}\)+ \(\frac{1}{11}\)= \(\frac{112}{11}\)

\(A=\frac{3}{2}-\frac{5}{6}+\frac{13}{12}-\frac{19}{20}+\frac{31}{30}-\frac{41}{42}+\frac{57}{56}-\frac{71}{72}+\frac{91}{90}-\frac{109}{110}\)

\(\Rightarrow A=\left(1+\frac{1}{2}\right)-\left(1-\frac{1}{6}\right)+\cdot\cdot\cdot+\left(1+\frac{1}{90}\right)-\left(1-\frac{1}{110}\right)\)

\(\Rightarrow A=1+\frac{1}{2}-1+\frac{1}{6}+\cdot\cdot\cdot+1+\frac{1}{90}-1+\frac{1}{110}\)

\(\Rightarrow A=\left[\left(1-1\right)+\frac{1}{2}+\frac{1}{6}\right]+\cdot\cdot\cdot+\left[\left(1-1\right)+\frac{1}{90}+\frac{1}{110}\right]\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{90}+\frac{1}{110}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow A=1-\frac{1}{11}\)

\(\Rightarrow A=\frac{10}{11}\)

Kết quả là 8090 / 819

\(=1+1+1+1+1+1+1+1+1+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=9+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

=9+9/10=99/10

\(2x+1+\frac{1}{6}+1+\frac{1}{12}+..+1+\frac{1}{90}=10\)

=> 2x + 8 + \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=10\)

=> 2x + \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=10-8\)

\(2x+1-\frac{1}{10}=2\)

=> 2x + \(\frac{9}{10}=2\)

=> 2x          = 2 - 9/10

=>2x           = 11/10 

=> x              = 11/10 : 2

x                  =  11/20 

thang Tran ơi,2x+1-1/10 ở đâu vậy

Phải là 2x+1/2-1/10 chứ

mk giải cho nè nhớ tích nha

2x+7/6+13/12+21/20+31/30+43/42+57/56+73/72+91/90=10

2x+(1/6+1)+(1/12+1)+(1/20+1)+(1/30+1)+(1/42+1)+(1/56+1)+(1/72+1)+(1/90+1)=10

2x+(1+1+1+1+1+1+1+1)+(1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)=10

2x+8+(1/2+3+1/3+4+1/4+5+1/5+6+1/6+7+1/7+8+1/8+9+1/9+10)=10

2x+8+(1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)=10

2x+8+(1/2-1/10)=10

2x+8+2/5=10

2x=10-8-2/5

2x=1,6

x=1,6:2

x=0,8

Mk thấy đề bài hơi dư nên x mk tìm ra là số thập phân còn đổi lại ra phân số là 4/5