BC
1
K
Khách
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NC
3
TQ
7
24 tháng 5 2016
\(\frac{4}{1x3x5}+\frac{4}{3x5x7}+...+\frac{4}{9x11x13}\)
\(=\frac{1}{1x3}-\frac{1}{3x5}+\frac{1}{3x5}-...+\frac{1}{9x11}-\frac{1}{11x13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
HL
1
OY
12 tháng 8 2021
\(\dfrac{4}{1\times3\times5}+\dfrac{4}{3\times5\times7}+\dfrac{4}{5\times7\times9}+\dfrac{4}{7\times9\times11}\)
=\(\dfrac{5-1}{1\times3\times5}+\dfrac{7-3}{3\times5\times7}+\dfrac{9-5}{5\times7\times9}+\dfrac{11-7}{7\times9\times11}\)
=\(\dfrac{1}{1\times3}-\dfrac{1}{3\times5}+\dfrac{1}{3\times5}-\dfrac{1}{5\times7}+...+\dfrac{1}{9\times11}-\dfrac{1}{11\times13}\)
=\(\dfrac{1}{3}-\dfrac{1}{143}=\dfrac{140}{429}\)
DN
2
PT
4
6 tháng 4 2018
\(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)
= \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+\frac{1}{7.9.11}+\frac{1}{9.11.13}\)
= \(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
= \(\frac{1}{1.3}-\frac{1}{11.13}\)
= \(\frac{140}{429}\)
~~~
Không chắc chắn lắm nhé :3
#Sunrise
HM
6 tháng 4 2018
nhớ cho k nhé !
4/1x3x5 = 1/1x3 - 1/3x5
4/3x5x7 = 1/3x5 - 1/5x7
.............
A = 1/1x3 - 1/11x13
1/1x3x5 = 1/4 x (1/1x3 - 1/3x5)
1/3x5x7 = 1/4 x (1/3x5 - 1/5x7)
..........
B = 1/4 x (1/1x3 - 1/11x13)
NG
1
TH
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 4
A = \(\dfrac{2}{1\times3\times5}\) + \(\dfrac{2}{3\times5\times7}\) + \(\dfrac{2}{5\times7\times9}\)+\(\dfrac{2}{7\times9\times11}\)
A = \(\dfrac{1}{2}\) x (\(\dfrac{4}{1\times3\times5}\) + \(\dfrac{4}{3\times5\times7}\) + \(\dfrac{4}{5\times7\times9}\) + \(\dfrac{4}{7\times9\times11}\))
A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\)-\(\dfrac{1}{3\times5}\)+\(\dfrac{1}{3\times5}\)-\(\dfrac{1}{5\times7}\)+\(\dfrac{1}{5\times7}\)-\(\dfrac{1}{7\times9}\)+\(\dfrac{1}{7\times9}\)-\(\dfrac{1}{9\times11}\))
A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\) - \(\dfrac{1}{9\times11}\))
A = \(\dfrac{1}{2}\) x (\(\dfrac{1}{3}-\dfrac{1}{99}\))
A = \(\dfrac{1}{2}\times\) \(\dfrac{32}{99}\)
A = \(\dfrac{16}{99}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 4
B = \(\dfrac{1}{1\times2\times3}\) + \(\dfrac{1}{2\times3\times4}\) + \(\dfrac{1}{3\times4\times5}\) + \(\dfrac{1}{4\times5\times6}\)
B = \(\dfrac{1}{2}\) x (\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+\dfrac{2}{4\times5\times6}\))
B = \(\dfrac{1}{2}\) x (\(\dfrac{1}{1\times2}\)-\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{2\times3}\)-\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{3\times4}\)-\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{4\times5}\)-\(\dfrac{1}{5\times6}\))
B = \(\dfrac{1}{2}\)x(\(\dfrac{1}{1\times2}\) - \(\dfrac{1}{5\times6}\))
B = \(\dfrac{1}{2}\)x (\(\dfrac{1}{2}-\dfrac{1}{30}\))
B = \(\dfrac{1}{2}\)x \(\dfrac{7}{15}\)
B = \(\dfrac{7}{30}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 4
A = \(\dfrac{4}{1\times3\times5}\) + \(\dfrac{4}{3\times5\times7}\) +\(\dfrac{4}{5\times7\times9}\) + \(\dfrac{4}{7\times9\times11}\) + \(\dfrac{4}{9\times11\times13}\)
A = \(\dfrac{1}{1\times3}\)-\(\dfrac{1}{3\times5}\)+\(\dfrac{1}{3\times5}\)-\(\dfrac{1}{5\times7}\)+...+\(\dfrac{1}{9\times11}\)-\(\dfrac{1}{11\times13}\)
A = \(\dfrac{1}{1\times3}\) - \(\dfrac{1}{11\times13}\)
A = \(\dfrac{1}{3}-\dfrac{1}{143}\)
A = \(\dfrac{140}{429}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 4
Bài 2:
A = \(\dfrac{1991}{1990}\) x \(\dfrac{1992}{1991}\) x \(\dfrac{1993}{1992}\) x \(\dfrac{1994}{1993}\) x \(\dfrac{1995}{997}\)
A = \(\dfrac{1994\times1995}{1990\times997}\)
A = \(\dfrac{997\times2\times5\times399}{5\times2\times199\times997}\)
A = \(\dfrac{399}{199}\)
\(A=9\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\)
\(=9\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{25\cdot27}-\dfrac{1}{27\cdot29}\right)\)
\(=9\left(\dfrac{1}{3}-\dfrac{1}{783}\right)=\dfrac{260}{87}\)