2/1.3 + 2/3.5 + 2/5.7 + ... + 2/x. (x + 2) =19
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TM
1
16 tháng 9 2020
Sửa đề: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2020}{2021}\) \(Đkxđ:\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2020}{2021}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{2020}{2021}\)
\(\Leftrightarrow\frac{x+2}{2021}=1\)
\(\Leftrightarrow x=2019\)
Vậy \(x=2019\)
NN
0
TM
0
VT
0
29 tháng 7 2023
=>2/1*3+2/3*5+...+2/(2x-1)(2x+1)=98/99
=>1-1/3+1/3-1/5+...+1/(2x-1)-1/(2x+1)=98/99
=>1-1/(2x+1)=98/99
=>1/(2x+1)=1/99
=>2x+1=99
=>x=49
NT
9
28 tháng 4 2015
a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
HD
1
TT
14 tháng 4 2016
a.2/1.3+2/3.5+2/5.7+................+2/99.101
1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101
1-1/101
100/101
b.5/1.3+5/3.5+5/5.7+............+5/99.101
5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2
5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)
5/2(1-1/3+1/3-1/5+1/5-1/7+........+1/99-1/101)
5/2(1-1/101)
5/2.100/101
250/101
H
15 tháng 5 2017
Có:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{5}{11}\)
\(\Rightarrow\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)
\(\Rightarrow\dfrac{1}{2}.\left(1-0-0-0...-0-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)
\(\Rightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)
\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{5}{11}:\dfrac{1}{2}=\dfrac{10}{11}\)
\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{10}{11}\)
\(\Rightarrow\dfrac{1}{x+2}=\dfrac{1}{11}\)
\(\Rightarrow x+2=11\)
\(\Rightarrow x=11-2=9\)
Vậy x = 9.
Chúc bạn học tốt!
15 tháng 5 2017
1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2)
= 1/2.(1/1 - 1/3) + 1/2.(1/3 - 1/5) + 1/2.(1/5 - 1/7) + ... + 1/2.(1/x -1/x+2)
= 1/2.(1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2 )
= 1/2.(1/1 - 0 - 1/x+2 )
= 1/2 . ( 1/1 - 1/x+2 )
= 1/2 . ( x+2/x+2 - 1/x+2 )
= 1/2 . x+1/x+2
Mà 1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2) = 5/11
=> 1/2 . x+1/x+2 = 5/11
=> x+1/x+2 = 5/11 : 1/2
=> x+1/x+2 = 10/11
=> x+1/x+2-1 = 10/11-1
=> x+1/x+1 = 10/10
=> x + 1 = 10
=> x = 10 - 1
=> x = 9
Vậy x = 9
NK
3
LV
12 tháng 8 2016
Ta có:
1/1.3 + 1/3.5 + 1/5.7 + ... + 1/x.(x+2) = 1/2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/x.(x+2)
= 1/2.(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2
= 1/2.(1 - 1/x+2)
=> 1/2.(1 - 1/x+2) = 20/41
1 - 1/x+ 2 = 20/41 : 1/2
1 - 1/x+2 = 40/41
1/x+2 = 1/41
=>x + 2 = 41
=>x = 41 - 2
=>x = 39
Vậy x = 39
Ủng hộ nha
12 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
=> \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=2.\frac{20}{41}\)
=> \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)
=> \(1-\frac{1}{x+2}=\frac{40}{41}\)
=> \(\frac{1}{x+2}=1-\frac{40}{41}\)
=> \(\frac{1}{x+2}=\frac{1}{41}\)
=> \(x+2=41\)
=> \(x=41-2=39\)
Ta có : \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{x\left(x+2\right)}=19\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{x}-\frac{1}{x+2}=19\)
\(\Leftrightarrow1-\frac{1}{x+2}=19\)
\(\Leftrightarrow\frac{x+2}{x+2}-\frac{1}{x+2}=19\)
\(\Leftrightarrow\frac{x+1}{x+2}=19\)
<=> 19(x + 2) = x + 1
<=> 19x + 38 = x + 1
=> 19x - x = 1 - 38
=> 19x = -37
=> x = \(-\frac{37}{19}\)
ĐK: \(x\ne0;x\ne2\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=19\)
\(\Leftrightarrow\)\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=19\)
\(\Leftrightarrow\)\(1-\frac{1}{x+2}=19\)
\(\Leftrightarrow\)\(\frac{1}{x+2}=-18\)
\(\Rightarrow\)\(x+2=-\frac{1}{18}\)
\(\Leftrightarrow\)\(x=-2\frac{1}{18}\)