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Snow Princess
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 Mashiro Shiina
16 tháng 3 2018 lúc 20:04

\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(P=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)

\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)

\(P=50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)

\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)

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Cao Thùy Linh
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Cao Thùy Linh
27 tháng 4 2023 lúc 23:08

Giúp mình với mình đang cần gấp!!!

 

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Vũ Đào
27 tháng 4 2023 lúc 23:16

=> D + 49 = (1/49 + 1) + (2/48 + 1) +... (49/1 + 1)

= 50/1 + 50/2 + ... + 50/49

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

=> D = 50(1/2 + 1/3 +... + 1/49) + 1

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

=> C/D = 1/50

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Kang Daniel
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Nguyễn Chơn Nhân
5 tháng 5 2018 lúc 16:03

P = 1/49+2/48+3/47+...+48/2+49/1

Cộng 1 váo mỗi p/s trong 48 p/s đầu , trừ p/s cuối đi 48 ta được

P=(1/49+1)+(2/48+1)+...+(48/2+1)+1

P= 50/49+50/48+....+50/2+50/50

Đưa ps cuối lên đầu

P=50/50+50/49+50/48+...+50/2

=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)

=50S

=> S/P=1/50

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Kang Daniel
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Kang Daniel
4 tháng 5 2018 lúc 18:43

Giúp vớikhocroi

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Jeon Jungkook
18 tháng 7 2018 lúc 15:11

Ta có:

P= \(\dfrac{1}{49}+\dfrac{2}{48}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

P= \(\dfrac{1}{49}+\dfrac{2}{48}+...+\dfrac{48}{2}+\left(1+1+...+1\right)\)(có 49 chữ số 1)

P= \(\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)

P= \(\dfrac{50}{49}+\dfrac{50}{48}+...+\dfrac{50}{2}+\dfrac{50}{50}\)

P= \(50.\left(\dfrac{1}{50}+\dfrac{1}{49}+...+\dfrac{1}{2}\right)\)

\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}}{50.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)}\)

\(\dfrac{S}{P}=\dfrac{1}{50}\)

Vậy \(\dfrac{S}{P}=\dfrac{1}{50}\)

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Mori Rannnnnnnnnnnnnnnnn...
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Nguyen My Van
18 tháng 5 2022 lúc 15:42

\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)

\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)

\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)

\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)

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Alan Walker
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Mới vô
11 tháng 5 2017 lúc 18:12

\(Q=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{47}{3}+\dfrac{48}{2}+\dfrac{49}{1}\\ =\dfrac{1}{49}+1+\dfrac{2}{48}+1+\dfrac{3}{47}+1+...+\dfrac{47}{3}+1+\dfrac{48}{2}+1+1\\ =\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{3}+\dfrac{50}{2}+\dfrac{50}{50}\\ =50\cdot\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+...+\dfrac{1}{3}+\dfrac{1}{2}\right)\\ =50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)\)

\(\dfrac{P}{Q}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}}{50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)

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Phạm Vũ Thanh Nhàn
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Nguyễn Ngọc Minh Châu
24 tháng 3 2017 lúc 16:25

Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)

\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)

\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)

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Nguyễn Kiều Anh
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Nguyễn Lê Phước Thịnh
18 tháng 9 2021 lúc 21:44

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

 

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MC Alextion
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Phùng Khánh Linh
2 tháng 10 2018 lúc 18:09

\(VT=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{48}+\sqrt{49}}=\dfrac{\sqrt{2}-1}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}+\dfrac{\sqrt{4}-\sqrt{3}}{1}+...+\dfrac{\sqrt{49}-\sqrt{48}}{1}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{49}-\sqrt{48}=\sqrt{49}-1=7-1=6=VP\)Vậy , đẳng thức được CM

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