1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50

=1/1-1/2+1/3-1/4+...+1/49-1/50

=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)

=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)

=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25

=1/26+1/27+...+1/50 (đpcm)

bn ơi bn có thê

rhuowngs dẫn mình 

làm ko vì

mai mình ucngx

có bài này

1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50

=1/1-1/2+1/3-1/4+...+1/49-1/50

=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)

=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)

=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25

=1/26+1/27+...+1/50 (đpcm)

nhé !

Ta có : 

\(1>\frac{1}{10}=\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)

\(............\)

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

\(\Rightarrow\)\(A=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\)

Do từ \(1\) đến \(100\) có \(100-1+1=100\) số tự nhiên nên có \(100\) phân số \(\frac{1}{\sqrt{100}}\) ta được : 

\(A>100.\frac{1}{\sqrt{100}}=\frac{100}{\sqrt{100}}=\frac{100}{10}=10\)

\(\Rightarrow\)\(A>10\) ( đpcm ) 

Vậy \(A>10\)

Chúc bạn học tốt ~ 

A = 1/10 + 1/12 + 1/14 + ... + 1/20 > 1/2

   = 1/2.5 + 1/2.6 + 1/2.7 + ... + 1/2.10 > 1/2

   = 1/2 . 1/5 + 1/2 . 1/6 + 1/2 . 1/7 + ... + 1/2 . 1/10 > 1/2

   = 1/2 . ( 1/5 + 1/6 + 1/7 + ... + 1/10 ) > 1/2   => (đpcm)

3/10=3/9*10

3/11=3/10*11

3/12=3/11*12

3/13=3/12*13

3/14=3/13*14

suy ra 3/10+3/3/11+....+3/14 nhỏ hơn 3/9*10+....+3/13*14

suy ra 3/9*10 + 3/10*11+....+3/13*14

=1/9-1/10+....+1/13-1/14

=1/9-1/14

tự viết kết quả nhé

1/1²+2² + 1/2²+3² + 1/3²+4² + ... + 1/10²+11²

< 1/1²+1² + 1/2²+2² + 1/3²+3² + ... + 1/10²+10²

< 1/2.1² + 1/2.2² + 1/2.3² + ... + 1/2.10²

< 1/2.(1/1² + 1/2² + 1/3² + ... + 1/10²)

< 1/2.(1 + 1/1.2 + 1/2.3 + ... + 1/9.10)

< 1/2.(1 + 1 - 1/2 + 1/2 - 1/3 + ... + 1/9 - 1/10)

< 1/2.(2 - 1/10)

< 1/2.(20/10 - 1/10)

< 1/2.19/10

< 19/20

Hình như bn chép sai đề

Ta có: \(\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}.\)

\(=\frac{1}{5}+\left(\frac{1}{14}+\frac{1}{31}+\frac{1}{44}\right)+\left(\frac{1}{61}+\frac{1}{84}+\frac{1}{96}\right)\)

Ta thấy \(\frac{1}{14}< \frac{1}{12}\)

            \(\frac{1}{31}< \frac{1}{12}\)

            \(\frac{1}{44}< \frac{1}{12}\)

\(=>\frac{1}{14}+\frac{1}{31}+\frac{1}{44}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}\)

\(=>\frac{1}{14}+\frac{1}{31}+\frac{1}{44}< \frac{1}{12}.3\left(1\right)\)

Ta lại thấy \(\frac{1}{61}< \frac{1}{60}\)

                \(\frac{1}{84}< \frac{1}{60}\)

                \(\frac{1}{96}< \frac{1}{60}\)

\(=>\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}\)

\(=>\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{60}.3\left(2\right)\)

Từ (1) và (2) suy ra: \(\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)

\(=>\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{5}+3.\left(\frac{1}{12}+\frac{1}{60}\right)\)

\(=>\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{2}\)

\(=>Đpcm\)

Ta có:\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..........+\frac{1}{64}\)

=\(1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+.........+\left(\frac{1}{33}+......+\frac{1}{64}\right)\)

\(>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+...+\left(\frac{1}{64}+\frac{1}{64}+.........+\frac{1}{64}\right)\)

=\(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)

=4

Vậy \(1+\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{64}>4\)

giup minh ,minh h cho