so sánh : \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\)với 1
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8
Q
9 tháng 4 2017
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=1-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
9 tháng 4 2017
Ta có :
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...........+\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
~ Chúc bn học tốt ~
6 tháng 5 2023
Ta có:
1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50 = 1/26 + 1/27 + 1/28 + .. + 1/50
Xét vế trái:
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 + ... + 1/49 ) - ( 1/2 + 1/4 + 1/6 + ... + 1/50 )
= ( 1 + 1/3 + 1/5 + ... + 1/49 ) + (1/2 + 1/4 + 1/6 + ... + 1/50 ) - 2 . ( 1/2 + 1/4 + 1/6 + ... + 1/50 )
= ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/49 + 1/50 ) - ( 1 + 1/2 + 1/3 + ... + 1/25 )
= 1/26 + 1/27 + 1/28 + ... + 1/49 + 1/50 (1)
Từ (1) => Vế trái = Vế phải
=> Điều phải chứng minh
- HokTot -
7 tháng 10 2021
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =1+\dfrac{1}{2}+...+\dfrac{1}{50}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{25}\\ =\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
7 tháng 10 2021
tại sao \(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
AH
Akai Haruma
Giáo viên
12 tháng 12 2017
Lời giải:
\(\text{VT}=\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+....+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-\left(1+\frac{1}{2}+\frac{1}{3}....+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{49}+\frac{1}{50}\)
Do đó ta có đpcm.
23 tháng 6 2016
1/1.2 + 1/2.3 + ... + 1/49.50
Đặt A = 1/1.2 + 1/2.3 + ... + 1/49.50
A = 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/49 - 1/50
A = 1/1 - 1/50
A = 49/50
Vì 49/50 < 1
=> A < 1
23 tháng 6 2016
Ta có : 1/1.2 + 1/2.3 + ... + 1/49.50
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/49 - 1/50
= 1 - 1/50
= 49/50
Vì 49/50 > 1
Nên 1/1.2 + 1/2.3 + ... + 1/49.50 < 1
21 tháng 3 2017
\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
\(E=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
\(E=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{4949}{9900}\)
21 tháng 3 2017
\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{98.99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
...
25 tháng 10 2021
\(0,125.\dfrac{3}{7}-\dfrac{1}{8}.\dfrac{11}{7}=\dfrac{1}{8}.\dfrac{3}{7}-\dfrac{1}{8}.\dfrac{11}{7}=\dfrac{1}{8}\left(\dfrac{3}{7}-\dfrac{11}{7}\right)=\dfrac{1}{8}.-\dfrac{8}{7}=-\dfrac{1}{7}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}< 1\)
11.2+12.3+...+149.5011.2+12.3+...+149.50
=1−12+12−13+...+149−150=1−12+12−13+...+149−150
=1−150<1