Đặt S=\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\)

Ta thấy S có 40 số hạng

ta có:

S=\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\)=\(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}\right)\)

\(+\left(\frac{1}{71}+\frac{1}{72}+...+\frac{1}{80}\right)\)(mỗi 1 nhóm có 100 số hạng)

>\(\left(\frac{1}{50}+...+\frac{1}{50}\right)+\left(\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{70}+...+\frac{1}{70}\right)+\left(\frac{1}{80}+...+\frac{1}{80}\right)\)(mỗi 1 nhóm có 10 số hạng)

=\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\)=\(\frac{533}{840}\)>\(\frac{490}{840}\)=\(\frac{7}{12}\)

vậy S>\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\)(đpcm)

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 )

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.

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)

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

a) \(\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)\)

\(=\frac{31}{23}-\frac{15}{23}\)

\(=\frac{16}{23}\)

b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)

\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)

\(=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)

\(=\frac{1}{3}-1+1\)

\(=\frac{1}{3}\)

c) \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{52}-\frac{3}{11}\right)\)

\(=\frac{38}{45}-\frac{8}{45}+\frac{17}{52}+\frac{3}{11}\)

\(=\frac{30}{45}+\frac{17}{52}+\frac{3}{11}\)

\(=\frac{2}{3}+\frac{17}{52}+\frac{3}{11}\)

\(=\frac{104+51}{156}+\frac{3}{11}\)

\(=\frac{155}{156}+\frac{3}{11}\)

\(=\frac{156}{156}-\frac{1}{156}+\frac{3}{11}\)

\(=1-\frac{1}{156}+\frac{3}{11}\)

\(=1-\left(\frac{11-468}{1716}\right)\)

\(=1-\frac{-457}{1716}\)

\(=1+\frac{457}{1716}\)

\(=\frac{2173}{1716}\)

a)31/23-(7/32+8/23)=31/23-7/32-8/23=(31/23-8/23)-7/32=1-7/32=25/32