\(S=\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)\)

\(S>\frac{50.1}{150}+\frac{50.1}{200}\)

\(\Rightarrow S>\frac{1}{3}+\frac{1}{4}\)

\(S>\frac{7}{12}\)

Chúc em học tốt^^

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}>\frac{7}{12}\)

\(\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)\)

\(\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)>\frac{50}{150}+\frac{50}{200}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{200}.\)

mik sẽ làm theo cách ngắn nhất mak cô đã bày :3 sai thì bạn ib mik để mik sửa ạ 

ta có \(\frac{1}{101}>\frac{1}{200}\)

      \(\frac{1}{102}>\frac{1}{200}\)

\(\frac{1}{103}>\frac{1}{200}\)

tương tự như vậy .... cho đến 

\(\frac{1}{199}>\frac{1}{200}\)

mak t có \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{200}.\)có 100 phân số

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{200}>100\cdot\frac{1}{200}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{200}>\frac{100}{200}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{200}>\frac{1}{2}\)(đpcm)

i

help me

a/  Tinh giá trị:

\(D=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{10}\right)\) \(\Leftrightarrow D=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{7}{8}.\frac{8}{9}.\frac{9}{10}=\frac{1}{10}\) 

b/  Chứng minh:

\(E=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\) 

-  Với mọi số tự nhiên n khác không thì luôn có:   \(\frac{1}{n^2}< \frac{1}{\left(n-1\right)\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\) Do đó:

 \(E=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}=\) 

   \(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\right)\)\(=\frac{1}{2}\left(1-\frac{1}{101}\right)< \frac{1}{2}\) Vậy \(E< \frac{1}{2}\) 

c/  Chứng minh : \(F=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{199}+\frac{1}{200}>\frac{7}{12}\) 

    \(F=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)>\frac{50}{150}+\frac{50}{200}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

   Vậy:            \(F>\frac{7}{12}\) .

1/100² + 1/101² + ... + 1/199² < 1/99.100 + 1/100.101 + ... + 1/198.199 = 1/99 - 1/100 + 1/100 - 1/101 + ... + 1/198 - 1/199 = 1/99 - 1/199

\(\Rightarrow\)Vậy 1/100² + 1/101² + ... + 1/199² < 1/99 (vì 1/99 đã lớn hơn 1/99 - 1/199 rồi mà G lại còn bé hơn 1/99 - 1/199 nữa)

1/100² + 1/101² + ... + 1/199² > 1/100.101 + ... + 1/199.200 = 1/100 - 1/101 + ... + 1/199 - 1/200 = 1/100 - 1/200 = 1/200

\(\Rightarrow\)Vậy 1/100² + 1/101² + ... + 1/199²  > 1/200

Ta có: \(\dfrac{1}{101}>\dfrac{1}{200};\dfrac{1}{102}>\dfrac{1}{200};....;\dfrac{1}{199}>\dfrac{1}{200}\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)

Vậy...

b