E=((2/2.3-1).(2/3.4-1).(2/4.5-1).....(2/2019.2020-1)
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C
3 tháng 5 2022
\(\text{#}HaimeeOkk\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}\)
\(A=1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{2019}-\dfrac{1}{2019}\right)-\dfrac{1}{2020}\)
\(A=1-0-0-0-...-0-\dfrac{1}{2020}\)
\(A=1-\dfrac{1}{2020}\)
\(A=\dfrac{2019}{2020}\)
Vậy \(A=\dfrac{2019}{2020}\)
PM
0
15 tháng 1 2017
1+2.( 1/2-1/3+1/3-1/4+....+1/(x-1)-1/x+1)=3/2
1+2.(1/2-1/x+1)=3/2
1-2/x+1=3/2-1
tự tính
9 tháng 4 2020
a) Ta có:
1/( 2.3 ) = ( 3 - 2 )/( 2.3 )
= 3/( 2.3 ) - 2/( 2.3 )
= 1/2 - 1/3.
1/( 3.4 ) = ( 4 - 3 )/( 3.4 )
= 4/( 3.4 ) - 3/( 3.4 )
= 1/3 - 1/4.
b)
Ta có:
A = 1/( 5.6 ) + 1/( 6.7 ) + 1/( 7.8 ) + ..... + 1/( 2019.2020 )
A = 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ..... + 1/2019 - 1/2020
A = 1/5 - 1/2020
A = 403/2020
Vậy A = 403/2020.
NN
9 tháng 4 2020
a) Ta có: \(\frac{1}{2.3}=\frac{3-2}{2.3}=\frac{3}{2.3}-\frac{2}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3.4}=\frac{4-3}{3.4}=\frac{4}{3.4}-\frac{3}{3.4}=\frac{1}{3}-\frac{1}{4}\)
b) Ta có: \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.......+\frac{1}{2019.2020}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+........+\frac{1}{2019}-\frac{1}{2020}\)
\(=\frac{1}{5}-\frac{1}{2020}=\frac{403}{2020}\)
X
9 tháng 2 2021
ĐKXĐ: \(x\ne0;x\ne-1\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2008}{2010}\)
\(\Leftrightarrow2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2008}{2010}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)(Biết công thức này chứ?)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)
\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2008}{2010}\)
\(\Leftrightarrow\dfrac{x-1}{x+1}=\dfrac{2008}{2010}\Leftrightarrow2010x-2010=2008x+2008\Leftrightarrow x=2009\left(tm\right)\)
Vậy x = 2009
24 tháng 6 2019
\(Q=1+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{50\cdot51}\)
\(Q=1+2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{50\cdot51}\right)\)
\(Q=1+2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(Q=1+2\cdot\left(\frac{1}{2}-\frac{1}{51}\right)\)
\(Q=1+\frac{49}{51}\)
\(Q=\frac{100}{51}\)
