LT
1
K
Khách
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E
1
Y
21 tháng 6 2019
+ \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{399\cdot400}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{400}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{400}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{400}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)\)
\(=\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{400}\)
+ \(\frac{1}{201\cdot400}+\frac{1}{202\cdot399}+...+\frac{1}{300\cdot301}\)
\(=\frac{1}{601}\cdot\left(\frac{201+400}{201\cdot400}+\frac{202+399}{202\cdot399}+...+\frac{300+301}{300\cdot301}\right)\)
\(=\frac{1}{601}\cdot\left(\frac{1}{201}+\frac{1}{400}+\frac{1}{202}+\frac{1}{399}+...+\frac{1}{300}+\frac{1}{301}\right)\)
\(=\frac{1}{601}\left(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{400}\right)\)
Do đó : \(A=\frac{1}{\frac{1}{601}}=601\)
DD
Đoàn Đức Hà
Giáo viên
12 tháng 7 2021
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(B=1.2+2.3+3.4+...+49.50\)
\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(=49.50.51\)
\(B=\frac{49.50.51}{3}=49.50.17\)
\(50^2.A-\frac{B}{17}=49.50-49.50=0\)
NN
5
LC
19 tháng 6 2015
Ta thấy:\(\frac{1}{1.2}=1-\frac{1}{2},\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3},...,\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
=>\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
=>\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
=>\(A=1-\frac{1}{50}\)
=>\(A=\frac{49}{50}\)
6 tháng 3 2018
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=1-\frac{1}{50}\)
\(\Rightarrow A=\frac{49}{50}\)
21 tháng 4 2022
Answer:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{100}{100}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
DQ
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 4 2023
A = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}...-\dfrac{1}{97.98}\)
A= 1-\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{97.98}\right)\)
A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)
A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{98}\right)\)
A=1- 1 + \(\dfrac{1}{98}\)
A= \(\dfrac{1}{98}\)
AH
Akai Haruma
Giáo viên
27 tháng 4 2023
Lời giải:
$1-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{97.98}$
$1-A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{98-97}{97.98}$
$1-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}$
$=1-\frac{1}{98}$
$\Rightarrow A=\frac{1}{98}$
NA
2 tháng 3 2022
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\\ =1-\dfrac{1}{2014}\\ =\dfrac{2013}{2014}\)
NT
0
10 tháng 4 2015
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
=1/1+-1/2+1/2+-1/3+1/3+-1/4+1/4+-1/5+1/5+-1/6
=1/1+-1/6=5/6
TK
0
\(S=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+....+\frac{1}{399\cdot400}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{399}-\frac{1}{400}\)
\(S=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+....+\frac{1}{399}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{400}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{400}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+.....+\frac{1}{400}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{400}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}\right)\)
\(S=\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{400}\)
P/S:Mik ms phân tích dc cái tử như thế này,còn mẫu thì mik phân tích dc nhưng A lại ko gọn cho lắm.