tính
\(\frac{1}{4.9}+\frac{1}{9.14}+.....
+\frac{1}{64.69}\)
K
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TQ
1
Những câu hỏi liên quan
20 tháng 8 2018
5B=\(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+...+\frac{5}{64\cdot69}\)
5B=\(\frac{9-4}{4\cdot9}+\frac{14-9}{9\cdot14}+...+\frac{69-64}{64.69}\)
5B=\(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\)
5B=\(\frac{65}{276}\)
B=\(\frac{13}{276}\)
KD
20 tháng 8 2018
\(B=\frac{1}{4.9}+\frac{1}{9.14}+....+\frac{1}{64.69}\)
\(\Rightarrow5B=\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{64.69}\)
\(5B=\frac{9-4}{4.9}+\frac{14-9}{9.14}+....+\frac{69-64}{64.69}\)
\(5B=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\)
\(5B=\frac{1}{4}-\frac{1}{69}\)
\(5B=\frac{65}{276}\)
\(B=\frac{65}{276}:5\)
\(B=\frac{13}{276}\)
29 tháng 1 2016
a 2^2015>3^1029
b 5A=\(\frac{5}{4.9}\)+\(\frac{5}{9.14}\)+\(\frac{5}{14.19}\)+.....+\(\frac{5}{64.69}\)
5A=1/4-1/9+1/9-1/14+1/14-1/19+1/19+....+1/64-1/69
5A=1/4-1/9
A=(1/4-1/9)/5
A=1/36
LS
9 tháng 7 2018
\(D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2017}{2018}\)
\(D=\frac{1}{2018}\)
Vậy \(D=\frac{1}{2018}\)
\(E=\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{64.69}+\frac{1}{69.74}\)
\(E=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{69}-\frac{1}{74}\right)\)
\(E=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{74}\right)\)
\(E=\frac{1}{5}\cdot\frac{35}{148}=\frac{7}{148}\)
Vậy E = ...
DP
23 tháng 7 2017
\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-....-49}{89}\)
\(\text{Đặt }:\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)\)là \(A\)
\(\frac{1-3-5-7-...-49}{89}\)là \(B\);ta có :
\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)
\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}\cdot\frac{45}{196}=\frac{9}{196}\)
\(B=\frac{1-3-5-7-....-49}{89}=\frac{1-\left(3+5+7+...+49\right)}{89}\)
Tổng của \(3+5+7+...+49\)là:
\(\frac{\left(3+49\right).24}{2}=624\)
\(\Rightarrow\frac{1-624}{89}=\frac{-623}{89}=-7\)
\(\Rightarrow\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-...-49}{89}=A.B=\frac{9}{196}\cdot-7=-\frac{9}{28}\)
25 tháng 3 2018
mk ko viết lại đề đâu
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)\(.\frac{1-\left(3+5+...+49\right)}{89}\)
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{\left(1-\frac{\left(49+3\right).24}{2}\right)}{89}\)
=\(\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\frac{52.24}{2}\right)}{89}\)
=\(\frac{9}{196}.\left(1-\frac{624}{89}\right)=\frac{9}{196}.\left(\frac{-623}{89}\right)\)
=\(\frac{-9}{28}\)
30 tháng 12 2017
ta có
1/5(5/36+5/126+...+5/44*49)1-3-5-7-9-...-49/89
=1/5(1/4-1/9+1/9-1/14+...+1/44-1/49)-623/89
=1/5*-7(1/4-1/49)
=-7/5*45/196
=-9/128
VA
1
25 tháng 3 2017
= \(\frac{1}{4}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{14}\)+\(\frac{1}{14}\)-\(\frac{1}{19}\)+... + \(\frac{1}{44}\)-\(\frac{1}{49}\)
= \(\frac{1}{4}\)-\(\frac{1}{49}\)
= \(\frac{45}{196}\)
ai tốt bụng thì tk cho mk nha, mk đg âm điểm nè huhu
2 tháng 10 2020
b) \(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{2-\left(1+3+5+7+..+49\right)}{12}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{2-\left(12.50+25\right)}{89}=-\frac{5.9.7.89}{5.4.7.7.89}=\frac{-9}{28}\)
1-1/64.69=6369/6469