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TRAN ANH BACH
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Long Vũ
4 tháng 4 2016 lúc 21:12

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

=>\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

=>\(A=1-\frac{1}{50}=\frac{49}{50}\)

mà A=49/50 

=>1/26+1/27+...+1/50 =49/50

Trần Thanh Quân
4 tháng 4 2016 lúc 21:36

49/50 ban oi

phạm hồng hạnh
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fan FA
24 tháng 8 2016 lúc 12:59

ta có:

1/1.2+1/3.4+1/5.6+...+1/49.50

=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50

=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)

=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2

=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)

=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50

hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50

Bui Nhat Minh
10 tháng 10 2017 lúc 16:32

bài này dễ ợt mình không làm đau

lê thị trà giang
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Uzumaki Naruto
3 tháng 9 2016 lúc 9:14

a)A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) > 1 / (1*2) + 1 / (3*4) = 1 / 2 + 1 / 12 = 7 / 12 ♦ 
A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100) = 
(1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 < 
1 - 1 / 2 + 1 / 3 = 5 / 6 ♥ 
♦, ♥ => 7 / 12 < A < 5 / 6

b)ta có:

1/1.2+1/3.4+1/5.6+...+1/49.50

=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50

=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)

=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2

=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)

=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50

hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50

Đặng Ngọc Thảo Nguyên
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 ♫ Love Music  ♫
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Nguyễn Thanh Hằng
12 tháng 9 2017 lúc 17:06

Ta có :

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+.........+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.........+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+......+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+.......+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+.......+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+.....+\dfrac{1}{25}\right)\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+......+\dfrac{1}{50}\)

Vậy ...

 Mashiro Shiina
12 tháng 9 2017 lúc 17:03

Đặt:

\(PHUCDZ=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)

\(PHUCDZ=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(PHUCDZ=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)\)

\(PHUCDZ=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)

\(PHUCDZ=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)\)

\(PHUCDZ=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)

Đặt \(PHUCMAXDZ=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)

\(PHUCDZ=PHUCMAXDZ\) vậy ta có \(đpcm\)

Heartilia Hương Trần
12 tháng 9 2017 lúc 17:08

Ta có : \(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\) + \(\dfrac{1}{5.6}\) +...+ \(\dfrac{1}{49.50}\)

= (1-\(\dfrac{1}{2}\)) + (\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)) + (\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)) + ... + (\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\))

= (1+\(\dfrac{1}{3}\) +\(\dfrac{1}{5}\)+....+\(\dfrac{1}{49}\)) - ( \(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{6}\)+...+\(\dfrac{1}{50}\))

=(1+\(\dfrac{1}{3}\)+\(\dfrac{1}{5}\)+...+\(\dfrac{1}{49}\)) - 2(\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{6}\)+...+\(\dfrac{1}{50}\))

= (1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+...+\(\dfrac{1}{50}\)) - (1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+...+\(\dfrac{1}{25}\))

=\(\dfrac{1}{26}\)+\(\dfrac{1}{27}\)+...+\(\dfrac{1}{50}\) (đpcm)

phung thi thu huyen
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Nguyễn Hoài Thương
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Trương Nguyễn Anh Kiệt
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Trần Thanh Phương
4 tháng 11 2018 lúc 11:50

\(\frac{1}{1\cdot2}+\frac{1}{3+4}+...+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

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

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\left(đpcm\right)\)

Jack Yasuo
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Nguyễn Hưng Phát
3 tháng 7 2016 lúc 20:44

\(\frac{1}{1.2}+\frac{1}{3.4}+........+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+......+\frac{1}{50}-\left(1+\frac{1}{2}+....+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+........+\frac{1}{50}\)

\(\Rightarrowđpcm\)

Khải Nhi
3 tháng 7 2016 lúc 20:46

ta có:1/1.2+1/3.4+1/5.6+...+1/49.50=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50

Bấm mình nha