DN
1
K
Khách
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DG
1
S
23 tháng 9 2023
\(E=\dfrac{1}{1\times2}+\dfrac{2}{2\times4}+\dfrac{3}{4\times7}+\dfrac{4}{7\times11}+\dfrac{5}{11\times16}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}\)
\(=1-\dfrac{1}{16}=\dfrac{15}{16}\)
#kễnh
VT
1
28 tháng 8 2023
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{56}-\dfrac{1}{67}\)
\(=1-\dfrac{1}{67}=\dfrac{66}{67}\)
NB
2 tháng 3 2023
1/1x2 + 2/2x4 + 3/4x7 + 4/7x11 +...+ 8/29x37 + 9/37x46
=2-1/1x2+4-2/2x4+...+46-37/37x46
=1-1/2+1/2-1/4+...+1/37-1/46
=1-1/46
=45/46
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 6 2023
A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)
A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)
A = \(\dfrac{105}{106}\)
B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)
C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))
C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)
C = \(\dfrac{5}{51}\)
D = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)
D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)
D = \(\dfrac{8}{9}\)
E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))
E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)
E = \(\dfrac{147}{200}\)
NT
0
8 tháng 6 2018
a) \(\frac{2}{11x16}+\frac{2}{16x21}+...+\frac{2}{61x66}\)
\(=\frac{2}{5}x\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\right)\)
\(=\frac{2}{5}x\left(\frac{1}{11}-\frac{1}{66}\right)\)
\(=\frac{2}{5}x\frac{5}{66}\)
\(=\frac{1}{33}\)
b) \(\frac{2}{5x7}+\frac{4}{7x11}+\frac{3}{11x14}+\frac{4}{14x18}+\frac{5}{18x23}+\frac{7}{23x30}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(=\frac{1}{5}-\frac{1}{30}\)
\(=\frac{1}{6}\)
IA
8 tháng 6 2018
a, \(\frac{2}{11\times16}+\frac{2}{16\times21}+...+\frac{2}{61\times66}\)
\(=\frac{2}{5}\times\left(\frac{5}{11\times16}+...+\frac{5}{61\times66}\right)\)
\(=\frac{2}{5}\times\left(\frac{1}{11}-\frac{1}{16}+...+\frac{1}{61}-\frac{1}{66}\right)\)
\(=\frac{2}{5}\times\left(\frac{1}{11}-\frac{1}{66}\right)\)
\(=\frac{2}{5}\times\frac{5}{66}\)
\(=\frac{1}{33}\)
Vậy giá trị của biểu thức trên là : \(\frac{1}{33}\)
b,\(\frac{2}{5\times7}+\frac{4}{7\times11}+\frac{3}{11\times14}+\frac{4}{14\times18}+\frac{5}{18\times23}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}\)
\(=\frac{1}{5}-\frac{1}{23}\)
\(=\frac{18}{115}\)
Vậy giá trị của biểu thức trên là \(\frac{18}{115}\)
A=1/1-1/2+1/2-1/4+1/4-1/7+1/7-1/11+1/11-1/16+1/16-1/22+1/22-1/29
A=1/1-1/29
A=28/29