Thấy 1/41+1/42 +......+ 1/60 < 1/40 .20

     1/41 +1/42 + .....+1/60<1/2

mà 1/61 +1/62+......+1/80 < 1/60 .20 =1/3

suy ra 1/41+1/42+ .......+1/80 <1/2 +1/3=7/12(đpcm)

Lại có 1/41 +1/42 +.....+1/80 <1/40 .40 =1(đpcm)

Nhận xét : Từ \(\frac{1}{41}\rightarrow\frac{1}{80}\)có 40 phân số . Gọi tổng các phân số đó là A.Ta có thể nhóm các phân số thành hai nhóm rồi so sánh các phân số có tử giống nhau.

Ta có : \(A=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}\)

\(=\left[\frac{1}{41}+\frac{1}{42}+...+\frac{1}{59}+\frac{1}{60}\right]+\left[\frac{1}{61}+\frac{1}{62}+...+\frac{1}{79}+\frac{1}{80}\right]\)

Vì \(\frac{1}{41}>\frac{1}{42}>...>\frac{1}{60}>\frac{1}{61}>...>\frac{1}{80}\) nên \(A>\left[\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}+\frac{1}{60}\right]+\left[\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}+\frac{1}{80}\right]\)

\(A>\frac{20}{80}+\frac{20}{80}=\frac{1}{3}+\frac{1}{4}=\frac{4+3}{12}=\frac{7}{12}\)

Vậy : \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}>\frac{7}{12}\)

Ta có: 7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80 

1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80) 

Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60

=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60

và 1/61> 1/62> ... >1/79> 1/80

=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80

Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12

=> 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12

=> ĐPCM                      ( ĐPCM có nghĩa là điều phải chứng minh)

~ Học tốt ~ K cho mk nhé! Thank you.

#)Giải :

 \(A=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)

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

Ta có : \(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};...;\frac{1}{59}>\frac{1}{60}\)

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

Lại có : \(\frac{1}{61}>\frac{1}{81};\frac{1}{62};>\frac{1}{80};...;\frac{1}{79}>\frac{1}{80}\)

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

Cộng (1) và (2) ta được :

\(A>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\left(đpcm\right)\)

        #~Will~be~Pens~#

bn vào các câu hỏi tương tự là sẽ thấy mấy câu y chang câu của bn thôi

Ta có :

 \(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};\frac{1}{43}>\frac{1}{60};....;\frac{1}{60}=\frac{1}{60}\)

\(\Rightarrow\frac{1}{41}+\frac{1}{42}+....+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=20.\frac{1}{60}=\frac{1}{3}\)(1)

\(\frac{1}{61}>\frac{1}{80};\frac{1}{62}>\frac{1}{80};\frac{1}{63}>\frac{1}{80};....;\frac{1}{80}=\frac{1}{80}\)

\(\Rightarrow\frac{1}{61}+\frac{1}{62}+....+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+....+\frac{1}{80}=20.\frac{1}{80}=\frac{1}{4}\)(2)

Từ (1) và (2) \(\Rightarrow y=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+....+\frac{1}{80}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)(đpvm)

y=\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...........................+\frac{1}{79}+\frac{1}{80}\)(có 80 số hạng)

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

                 Có 20 số hạng                                                         Có 20 số hạng

\(>\left(\frac{1}{60}+\frac{1}{60}+................+\frac{1}{60}\right)+\left(\frac{1}{80}+\frac{1}{80}+..........+\frac{1}{80}\right)\)

                     Có 20 số hạng                                                Có 20 số hạng

=>\(y>20.\frac{1}{60}+20.\frac{1}{80}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

Vậy \(y>\frac{7}{12}\)

Chúc bn học tốt

Gọi \(B=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{60}\)

\(C=\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{80}\)

Ta có : \(B=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{60}>\frac{1}{60}.20=\frac{2}{3}\)

\(C=\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{80}>\frac{1}{80}.20=\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}>\frac{2}{3}+\frac{1}{4}=\frac{11}{12}\)

Mà \(\frac{11}{12}>\frac{7}{12}\Rightarrow\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}>\frac{7}{12}\)

khỉ thiệt 

uk

7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80
1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)
Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60
và 1/61> 1/62> ... >1/79> 1/80
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80
Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12
=> 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12

nhớ đúng cái

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}\)

\(=\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)>\frac{1}{60}.20+\frac{1}{80}.20\)

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

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

Ta có : 

\(\frac{7}{12}\)= \(\frac{4}{12}\)+ \(\frac{3}{12}\)= \(\frac{1}{3}\)+ \(\frac{1}{4}\)= \(\frac{20}{60}\)+ \(\frac{20}{80}\)

\(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ .... + \(\frac{1}{79}\)+ \(\frac{1}{80}\)= (\(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ ....+\(\frac{1}{60}\)) + ( \(\frac{1}{61}\)+ \(\frac{1}{62}\)+...+\(\frac{1}{79}\)+\(\frac{1}{80}\))

Do \(\frac{1}{41}\)>\(\frac{1}{42}\)>....>\(\frac{1}{60}\)

=> ( \(\frac{1}{41}\)+ \(\frac{1}{42}\)+...+\(\frac{1}{60}\)) > \(\frac{1}{60}\)+...+\(\frac{1}{60}\)= \(\frac{20}{60}\)

Vậy : \(\frac{1}{61}\)> \(\frac{1}{62}\)>....>\(\frac{1}{79}\)>\(\frac{1}{80}\)

=> ( \(\frac{1}{61}\)+\(\frac{1}{62}\)+...+\(\frac{1}{79}\)+ \(\frac{1}{80}\)) > \(\frac{1}{80}\)+...+ \(\frac{1}{80}\)= \(\frac{20}{80}\)

Vậy : \(\frac{1}{41}\)+ \(\frac{1}{42}\)+....+\(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{20}{60}\)+ \(\frac{20}{80}\)

Vậy : \(\frac{1}{41}\)+ \(\frac{1}{42}\)+....+ \(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{20}{60}\)+ \(\frac{20}{80}\)= \(\frac{7}{12}\)

=> ĐPCM