S
2
K
Khách
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A
0
8 tháng 10 2017
\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}.\frac{11}{10}\)
\(y=\frac{22}{30}\)
VV
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
29 tháng 3 2019
\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}\right)xX=\frac{8}{11}\)
\(\Rightarrow\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right)xX=\frac{16}{11}\)
\(\Rightarrow\left(\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}\right)xX=\frac{16}{11}\)
\(\Rightarrow\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)xX=\frac{16}{11}\)
\(\Rightarrow\left(\frac{1}{3}-\frac{1}{11}\right)xX=\frac{16}{11}\Rightarrow\frac{8}{33}xX=\frac{16}{11}\)
\(\Rightarrow X=\frac{16}{11}:\frac{8}{33}=\frac{16}{11}x\frac{33}{8}=6\)
T
5 tháng 8 2018
\(\frac{3}{5×3}+\frac{3}{5×7}+\frac{3}{7×9}+\frac{3}{9×11}+\frac{3}{11×13}\)
\(=\frac{3}{2}×\left(\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+\frac{2}{9×11}+\frac{2}{11×13}\right)\)
\(=\frac{3}{2}×\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{3}{2}×\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{3}{2}×\left(\frac{13}{39}-\frac{3}{39}\right)\)
\(=\frac{3}{2}×\frac{10}{39}\)
\(=\frac{5}{13}\)
KT
5 tháng 8 2018
\(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+\frac{3}{11.13}\)
\(=\frac{3}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{ 1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11} +\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{3}{2}.\frac{10}{39}\)
\(=\frac{15}{39}\)
TN
4
9 tháng 8 2018
\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)
\(A=\frac{1}{3}-\frac{1}{19}\)
\(A=\frac{16}{57}\)
Dấu "." là dấu nhân nhá ^^
L
3
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
13 tháng 1 2020
\(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{5}-\frac{1}{15}=\frac{3}{15}-\frac{1}{15}=\frac{3-1}{15}=\frac{2}{15}\)
P/s : Dấu chấm là nhân nhé!
LC
2
8 tháng 7 2016
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\frac{10}{11}\)
\(=\frac{5}{11}\)
IW
8 tháng 7 2016
\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{9\times11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\frac{10}{11}\)
\(=\frac{5}{11}\)
M
12 tháng 8 2018
p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)
<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019
<=> p = 1/3 - 1/2019
<=> p = 224/673
KT
12 tháng 8 2018
\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2019}\right)\)
\(=\frac{112}{673}\)
\(\frac{3}{5x7}+\frac{3}{7x9}+...+\frac{3}{59x61}\)
\(=\frac{3}{2}\left(\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{59x61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}++...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)
Nguyễn Tuấn Minh giải đúng rồi nhé