\(\frac{3}{5x7}+\frac{3}{7x9}+...+\frac{3}{59x61}\)
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NT
19 tháng 3 2018
Đặt \(A=\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{59.61}\)
\(\Rightarrow\frac{2}{3}.A=\frac{2}{3}.\left(\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{59.61}\right)\)
\(\Rightarrow\frac{2}{3}.A=\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{59.61}\)
\(\Rightarrow\frac{2}{3}.A=\frac{7-5}{5.7}+\frac{9-7}{7.9}+.....+\frac{61-59}{59.61}\)
\(\Rightarrow\frac{2}{3}.A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)
\(\Rightarrow\frac{2}{3}.A=\frac{1}{5}-\frac{1}{61}\)
\(\Rightarrow\frac{2}{3}.A=\frac{56}{305}\)
\(\Rightarrow A=\frac{56}{305}:\frac{2}{3}\)
\(\Rightarrow A=\frac{56}{305}.\frac{3}{2}\)
\(\Rightarrow A=\frac{84}{305}\)
Vậy \(\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{59.61}=\frac{84}{305}\)
15 tháng 4 2019
\(S=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(S=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.60}\right)\)
\(S=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{60}\right)\)
\(S=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{60}\right)\)
\(S=\frac{3}{2}.\left(\frac{12}{60}-\frac{1}{60}\right)\)
\(S=\frac{3}{2}.\frac{11}{60}\)
\(S=\frac{11}{40}\)
A
0
22 tháng 2 2022
\(\dfrac{6}{5.7}+\dfrac{6}{7.9}+...+\dfrac{6}{59.61}\)
\(=3\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
\(=3\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=3\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(=\dfrac{3.56}{305}\\ =\dfrac{168}{305}\)
8 tháng 10 2017
\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}.\frac{11}{10}\)
\(y=\frac{22}{30}\)
DP
12 tháng 7 2017
\(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+\frac{3}{9\cdot11}+...+\frac{3}{2013\cdot2015}\)
\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{2015}\right)=\frac{3}{2}\cdot\frac{402}{2015}-\frac{603}{2015}\)
Vậy \(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+\frac{3}{9\cdot11}+...+\frac{3}{2013\cdot2015}=\frac{603}{2015}\)
12 tháng 7 2017
3/5.7 + 3/7.9 + 3/9.11 + ... 3/2013.2015
= 3/2.( 2/5.7 + 2/7.9 + 2/9.11 + ... + 2/2013.2015)
= 3/2. ( 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/2013 - 1/2015)
~~~ SAU ĐÓ BẠN GẠCH ĐI NHỮNG PHÂN SỐ GIỐNG NHAU NHÁ ~~~
= 3/2. ( 1/5 - 1/2015)
= 3/2. 2010/10075
= 603/4030
Mk chắc chắn cách làm đúng đó!!!
VV
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
29 tháng 3 2019
\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}\right)xX=\frac{8}{11}\)
\(\Rightarrow\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right)xX=\frac{16}{11}\)
\(\Rightarrow\left(\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}\right)xX=\frac{16}{11}\)
\(\Rightarrow\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)xX=\frac{16}{11}\)
\(\Rightarrow\left(\frac{1}{3}-\frac{1}{11}\right)xX=\frac{16}{11}\Rightarrow\frac{8}{33}xX=\frac{16}{11}\)
\(\Rightarrow X=\frac{16}{11}:\frac{8}{33}=\frac{16}{11}x\frac{33}{8}=6\)
HA
20 tháng 4 2016
Q= 3/3x5 + 3/5x7 + 3/7x9 +...+ 3/47x49
Q= (3/3 -3/5) + (3/5-3/7) + (3/7-3/9)+...+(3/47-3/49)
Q= 3/3 - 3/5 + 3/5 - 3/7 + 3/7 - 3/9 + ... + 3/47 - 3/49
Q=3/3 - 3/49
Q= 46/49
\(\frac{3}{5x7}+\frac{3}{7x9}+...+\frac{3}{59x61}\)
\(=\frac{3}{2}\left(\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{59x61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}++...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)
Nguyễn Tuấn Minh giải đúng rồi nhé