A = \(\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{91}{90}+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+...+\left(1+\frac{1}{90}\right)\)

\(=\left(1+\frac{1}{1.2}\right)+\left(1+\frac{1}{2.3}\right)+\left(1+\frac{1}{3.4}\right)+...+\left(1+\frac{1}{9.10}\right)\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)+\left(1+1+1+...+1\right)\)(9 số hạng 1)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)+1.9\)

\(=\left(1-\frac{1}{10}\right)+9=10-\frac{1}{10}=\frac{99}{10}>\frac{98}{11}\)

a=1+1/1.2+1+1+1/2.3+....+1+1/9.10

a=1+1+...+1(9 chữ số 1)+1/1-1/2+1/2-1/3+...+1/9-1/10

a=9+1-1/10

a=9+9/10=9+0.9=9.9

b=98/11<98/10=9.8<9.9

=>vậy a>b

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 

Tham khảo

A=3/2+7/6+13/12+...+91/90

A=1+1/2+1+1/6+…+1+1/72+1+1/90

A=(1+1+1+…+1+1)+1/1.2+1/2.3+1/3.4+…+1/9.10

A=10+1/1-1/2+1/2-1/3+…-1/9+1/9+1/10

A=10+1-1/10

A=10+9/10

A=109/10

\(S=\dfrac{3}{2}+\dfrac{7}{6}+\dfrac{13}{12}+...+\dfrac{91}{90}\)

\(=1+\dfrac{1}{2}+1+\dfrac{1}{6}+1+\dfrac{1}{12}+...+1+\dfrac{1}{90}\)

\(=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right)+9\)

\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)+9\)

\(=1-\dfrac{1}{10}+9=\dfrac{99}{10}\)

a=1+1/1.2+1+1/2.3+....+1+1/9.10

a=1+1+...+1(9 chữ số 1) +1/1-1/2+1/2-1/3+..+1/9-1/10

a=9+1-1/10

a=9+9/10=9+0.9=9.9

b=98/11<98/10=9.8<9.9.

vậy a>b

Ta có: a=1+1/2+1+1/6+1+1/12+...+1+1/90=9+1/2+1/6+...+1/90 > 9>99/11> b. Vậy, a>b

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

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=1+1/2+1+1/6+1+1/12+...+1+1/90=

=9+1/2+1/6+1/12+...+1/90

1/2+1/6+1/12+...+1/90=

1/1x2+1/2x3+2/3x4+...+1/9x10=

\(=\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{10-9}{9x10}=\)

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

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

\(\Rightarrow A=9+\dfrac{9}{10}=9\dfrac{9}{10}\)