em nghĩ muốn đau đầu mà không nhớ
1/2+1/4+1/8+1/16+1/32
1/1x6+1/6x11+1/11x16....1/96x101
K
Khách
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Những câu hỏi liên quan
19 tháng 11 2017
Đặt \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{496.501}\)
\(\Rightarrow5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{496}+\frac{1}{501}\)
\(\Rightarrow5A=1-\frac{1}{501}=\frac{500}{501}\)
\(\Rightarrow A=\frac{500}{501}:5=\frac{500}{501}.\frac{1}{5}=\frac{100}{501}\)
19 tháng 11 2017
k mik nhé
=1/5x(1-1/6+1/6-1/11-1/16+...+1/496-1/501
=1/5x(1-1/501)
=1/5x500/501
=100/501
15 tháng 3 2019
\(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\)
\(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{101}\right)\)
\(=1.\frac{100}{101}\)
\(=\frac{100}{101}\)
Q
10 tháng 3 2017
\(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
10 tháng 3 2017
đặt biểu thức là A
5A=(1/1x6+1/6x11+...+1/96x101)x5=5/1x6+5/6x11+...+5/96x101
5A=6-1/1x6+11-6/6x11+...+101-96/96x101
5A=6/1x6-1/1x6+11/6x11-6/6x11+...+101/96x101-96/96x101
5A=1-1/6+1/6-1/11+...+1/96-1/101(sau khi rút gọn các phân số)
5A=1-1/101(còn lại sau khi trừ)
5A=100/101
A=100/101:5=20/101
7 tháng 5 2020
Lớp 4 đó nếu ai ko làm được thì ko phải học sinh giỏi đó nha
NK
4
NL
29 tháng 4 2016
Gọi A = 1/1.6 + 1/6.11 +...+ 1/(5n+1)(5n+6)
5A = 5/1.6 + 5/6.11 + ... + 5/(5n+1)(5n+6)
=1 - 1/6 + 1/6 - 1/11 + ... + 1/5n+1 - 1/5n+6
=1 - 1/5n+6 =5n+6/5n+6 - 1/5n+6=5n+5 /5n+6
QL
0
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 = \(\dfrac{147}{200}\)
a: Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
=>\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
=>\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\dfrac{1}{32}\)
=>\(A=1-\dfrac{1}{32}=\dfrac{31}{32}\)
b: Đặt \(B=\dfrac{1}{1\cdot6}+\dfrac{1}{6\cdot11}+...+\dfrac{1}{96\cdot101}\)
=>\(B=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{96\cdot101}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{5}\cdot\dfrac{100}{101}=\dfrac{20}{101}\)