1/3.5 +1/5.7+1/7.9+...+1/(2.x+1).(2.x+3)
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UN
1
TM
18 tháng 4 2017
<=> 2/1.3 + 2/3.5 + 2/5.7 +....+ 2/(x+2)(x+4) = 100/101
<=> 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +.....+ 1/x+2 - 1/x+4 = 100/101
<=> 1 - 1/x+4 = 100/101
<=> 1/x+4 = 1 - 100/101 <=> 1/x+4 = 1/101 <=> x+4 = 101 <=> x= 101 - 4 = 97
:)
11 tháng 7 2015
1/3.5+1/5.7+1/7.9+...+1/(2x+1)(2x+3)=5/31
1/2(2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3))=5/31
1/3-1/5+1/5-1/7+1/7-1/9+...+1/2x+1-1/2x+3=5/31:1/2
1/3-1/2x+3=10/31
1/2x+3=1/3-10/31
1/2x+3=1/63
suy ra : 2x+3=63
2x=63-3
2x=60
x=60:2
x=30
vay x=30
nhớ **** cho mình nha
TT
1
TT
1
HT
27 tháng 8 2015
\(\left[\frac{12}{11}-\left(\frac{1}{2}+\frac{1}{44}\right)\right].\left(x-0,2\right)=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{25}{44}.\left(x-0,2\right)=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{9.11}\right)\)
\(x-0,2=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right):\frac{25}{44}\)
\(x-\frac{1}{5}=\frac{22}{25}.\left(1-\frac{1}{11}\right)=\frac{22}{25}.\frac{10}{11}=\frac{4}{5}\)
\(x=\frac{4}{5}+\frac{1}{5}\)
\(x=1\)
NV
1
S
26 tháng 7 2016
\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)
\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)
\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)
\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)
SK
7
QT
1
18 tháng 4 2018
Đặt \(A=\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{x\left(x+2\right)}\)(sửa đề)
\(\Rightarrow A=\frac{1}{2}.3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(\Rightarrow A=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{x+2}\right)\)
\(\Rightarrow A=\frac{1}{2}-\frac{3}{2x+4}\)
28 tháng 4 2021
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1
11 tháng 5 2019
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)