\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{60}>\frac{1}{60}.\left(60-41+1\right)=\frac{1}{60}.20=\frac{1}{3}\)(1)

\(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{80}>\frac{1}{80}.\left(80-61+1\right)=\frac{1}{80}.20=\frac{1}{4}\)(2)

Từ (1)(2)=>\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\left(đpcm\right)\)

Ta có : \(\frac{7}{12}=\frac{4}{12}+\frac{3}{12}=\frac{1}{3}+\frac{1}{4}\)

Ta chia tổng S thành 2 tổng nhỏ hơn như sau :

\(S=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{79}+\frac{1}{80}=\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)

+) Vì \(\frac{1}{41}>\frac{1}{42}>\frac{1}{43}>...>\frac{1}{60}\) => \(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)

\(\Rightarrow\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)>\frac{1}{60}\times20\)

\(\Rightarrow\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)>\frac{1}{3}\)

+) Vì \(\frac{1}{61}>\frac{1}{62}>\frac{1}{63}>...>\frac{1}{80}\Rightarrow\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}\)

\(\Rightarrow\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)>\frac{1}{80}\times20\)

\(\Rightarrow\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)>\frac{1}{4}\)

\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{1}{3}+\frac{1}{4}\)

\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{7}{12}\)

Vậy \(S>\frac{7}{12}\) ( đpcm )

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Ta có:

\(\frac{1}{51}>\frac{1}{75}\)

\(\frac{1}{52}>\frac{1}{75}\)

......................

\(\frac{1}{75}=\frac{1}{75}\)

\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=25.\frac{1}{75}=\frac{1}{3}\)(1)

Ta có:

\(\frac{1}{76}>\frac{1}{100}\)

\(\frac{1}{77}>\frac{1}{100}\)

........................

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

\(\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=25.\frac{1}{100}=\frac{1}{4}\)(2)

Từ (1) và (2) ta có:

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{7}{12}\)(5)

Ta có:

\(\frac{1}{51}=\frac{1}{51}\)

\(\frac{1}{52}< \frac{1}{51}\)

...................

\(\frac{1}{75}< \frac{1}{51}\)

\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}=25.\frac{1}{51}< 25.\frac{1}{50}=\frac{1}{2}\)(3)

Ta có:

\(\frac{1}{76}=\frac{1}{76}\)

\(\frac{1}{77}< \frac{1}{76}\)

...................

\(\frac{1}{100}< \frac{1}{76}\)

\(\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}< \frac{1}{76}+\frac{1}{76}+...+\frac{1}{76}=25.\frac{1}{76}< 25.\frac{1}{75}=\frac{1}{3}\)(4)

Từ (3) và (4) ta có:

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{5}{6}\)(6)

Từ (5) và (6) 

\(\Rightarrow\frac{7}{12}< \frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}< \frac{5}{6}\)

                                                            đpcm

Tham khảo nhé~

mình vừa mới trả lời xong đấy 

Câu hỏi của Do Not Ask Why - Toán lớp 7 - Học toán với OnlineMath

Ta có :

A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

A = \(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

A = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

A = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

A =  \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

Tách A thành 2 nhóm,ta được :

A = \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

Lại có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75}\text{ }\text{ }\)

            \(\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\text{ }\text{ }\)

A > \(\left(\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)=\frac{1}{75}.25+\frac{1}{100}.25\)

\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

A < \(\left(\frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}\right)+\left(\frac{1}{76}+\frac{1}{76}+...+\frac{1}{76}\right)=\frac{1}{51}.25+\frac{1}{76}.25< \frac{1}{50}.25+\frac{1}{75}.25\)

\(=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(\frac{7}{12}< A< \frac{5}{6}\)

Bạn SKT_NTT làm đúng rồi nha

Số số hạng của A là:

(200-101):1+1=100(số)

Nếu ta nhóm A thành các nhóm,mỗi nhóm 50 số hạng ta được :

100:50=2(nhóm)

Ta có :

A=(1/101+1/102+...+1/150)+(1/151+1/152+1/153+...+1/200)

Vì 1/101<1/102<1/103<...<1/150 nên 1/101+1/102+...+1/150<1/150x50

1/151<1/152<1/153<...<1/200 nên 1/151+1/152+1/153+...+1/200<1/200x50

Từ 3 điều trên suy ra:

A<1/150x50+1/200x50

A<1/3+1/4

A<7/12

vậy A<7/12

❤~~~ HỌC TỐT~~~❤Đặng Khánh Duy

A:  có 30 số hạng không đủ 

phải chia nhỏ ra

\(A=\left(\frac{1}{31}+...+\frac{1}{36}\right)+\left(\frac{1}{37}+..+\frac{1}{48}\right)+\left(\frac{1}{49}+..+\frac{1}{60}\right)\)

\(A>\left(\frac{6}{36}\right)+\left(\frac{12}{48}\right)+\left(\frac{12}{60}\right)=\frac{3}{12}+\frac{3}{12}+\frac{1}{12}=\frac{7}{12}\)