2.4/1.5+6.8/5.9+10.12/9.13+...+42.44/41.45
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TH
3
N
29 tháng 7 2018
a) \(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{5}=\frac{2}{10}=\frac{1}{5}\)
b) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}\)
\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}\)
\(=1-\frac{1}{17}=\frac{16}{17}\)
hok tốt ...
BT
29 tháng 7 2018
a)\(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}\)
\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+\frac{2}{8\cdot10}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(A=\frac{2}{5}\cdot\frac{1}{2}=\frac{1}{5}\)
b)\(B=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}=1-\frac{1}{17}=\frac{16}{17}\)
T
3
KT
15 tháng 8 2018
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{44}{45}\)
\(=\frac{11}{45}\)
TM
15 tháng 8 2018
Đặt \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\) là A.
Ta có:
\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)
\(4A=4\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\right)\)
\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\)
\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\)
\(4A=1-\frac{1}{45}\)
\(4A=\frac{44}{45}\)
\(A=\frac{44}{45}:4\)
\(A=\frac{11}{45}\)
Vậy \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}=\frac{11}{45}\)
9 tháng 1 2022
\(\Leftrightarrow x+2\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}\right)=\dfrac{-74}{45}\)
\(\Leftrightarrow x+2\cdot\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=-\dfrac{74}{45}\)
hay x=-2
VD
23 tháng 4 2022
\(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=\dfrac{-37}{45}\\ \Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\\ \Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\\ \Leftrightarrow x+\dfrac{8}{45}=\dfrac{-37}{45}\\ \Leftrightarrow x=-1\)
15 tháng 8 2019
\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)
\(4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\)
\(4B=\frac{44}{45}\)
\(B=\frac{11}{45}\)
XO
15 tháng 8 2019
\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{41.45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{44}{45}\)
\(=\frac{11}{45}\)
VA
16 tháng 7 2016
Ta có :
\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\)\(=1\)
\(x+\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=1\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=1\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=1\)
\(x+\frac{8}{45}=1\)
\(x=1-\frac{8}{45}=\frac{37}{45}\)
Ủng hộ mk nha !!! ^_^
DN
0
PQ
5
LP
7 tháng 4 2015
Theo đề bài ta có
x+1/5-1/9+1/9-1/13+.........+1/41-1/45=-37/45
x+(1/5-1/45)=-37/45
x+8/45=-37/45
x=-37/45 - 8/45
x=-45/45
x=-1
NL
14 tháng 4 2017
Theo bài ra ta có:
x1/5-1/9+1/9-1/13+...+1/41-1/45=-37/45
x+8/45=-37/45
x=-45/45
x=-1/1
x=-1
NN
0
M
1