Tính giá trị biểu thức:
B= -1 - 1/3 - 1/6 - 1/10 - 1/15 - ... - 1/1225
K
Khách
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MA
1
H
5 tháng 4 2021
\(P=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+...+\left(1-\dfrac{1}{1225}\right)+\left(1-\dfrac{1}{1275}\right)\\ \Rightarrow\dfrac{P}{2}=\left(\dfrac{1}{2}-\dfrac{1}{6}\right)+\left(\dfrac{1}{2}-\dfrac{1}{12}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{2550}\right)\\ =\left(\dfrac{1}{2}-\dfrac{1}{2\cdot3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3\cdot4}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{50\cdot51}\right)\\ =\dfrac{1}{2}\cdot49-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\\ =\dfrac{49}{2}-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\\ =\dfrac{49}{2}-\dfrac{1}{2}+\dfrac{1}{51}=\dfrac{1225}{51}\\ \Rightarrow P=\dfrac{2450}{51}\)
ND
0
N
1
1 tháng 7 2016
B = 1/6 + 1/12 + 1/20 + ... + 1/90
B = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/9.10
B = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10
B = 1/2 - 1/10
B = 5/10 - 1/10
B = 4/10 = 2/5
Ủng hộ mk nha ♡_♡☆_☆
TK
3
1 tháng 6 2023
Ta có \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2021}\right)\left(1-\dfrac{1}{2022}\right)\)
\(B=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}.\dfrac{2021}{2022}\)
\(B=\dfrac{1}{2022}\)
NG
3
13 tháng 5 2022
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2020}{2021}\cdot\dfrac{2021}{2022}=\dfrac{1}{2022}\)
O
13 tháng 5 2022
\(B=\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot\cdot\cdot\left(1-\dfrac{1}{2021}\right)\cdot\left(1-\dfrac{1}{2022}\right)\)
\(B=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\cdot\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\cdot\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\cdot\cdot\cdot\left(\dfrac{2021}{2021}-\dfrac{1}{2021}\right)\cdot\left(\dfrac{2022}{2022}-\dfrac{1}{2022}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\cdot\cdot\dfrac{2020}{2021}\cdot\dfrac{2021}{2022}\)
\(B=\dfrac{1\cdot2\cdot3\cdot\cdot\cdot2020\cdot2021}{2\cdot3\cdot4\cdot\cdot\cdot2021\cdot2022}\)
\(B=\dfrac{1}{2022}\)
27 tháng 7 2021
b) Ta có: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}+1}-6\sqrt{\dfrac{16}{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1-6\cdot\dfrac{4}{\sqrt{3}}\)
\(=1-8\sqrt{3}\)
HQ
4
\(B=-1-\frac{1}{3}-\frac{1}{6}-...-\frac{1}{1225}\)
\(=-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=-2\left(1-\frac{1}{50}\right)=-2\cdot\frac{49}{50}=-\frac{49}{25}\)
\(B=-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)
\(B=-2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\right)\)
\(B=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(B=-2\left(1-\frac{1}{50}\right)\)
\(B=-2\cdot\frac{49}{50}\)
\(B=-\frac{49}{25}\)