E = \(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{972}\)
K
Khách
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Những câu hỏi liên quan
WW
4
21 tháng 6 2022
A=14+112+136+...+1972+12916
3A=34+14+112+...+1324+1972
3A−A=(34+14+112+...+1324+1972)−(14+112+136+...+1972+12916)
2A=34−12916
A=10932916
PA
1
26 tháng 3 2018
\(B=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\\\)
\(3B=3\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)
\(3B=\dfrac{3}{4}+\dfrac{3}{12}+\dfrac{3}{36}+\dfrac{3}{108}+\dfrac{3}{324}+\dfrac{3}{972}\)
\(3B=\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\)
\(2B=3B-B\)
\(2B=\left(\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\right)-\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)
\(2B=\dfrac{3}{4}-\dfrac{1}{972}=\dfrac{729-1}{972}=\dfrac{728}{972}=\dfrac{182}{243}\)
\(B=\dfrac{182}{243}:\dfrac{1}{2}=\dfrac{182\cdot2}{243}=\dfrac{364}{243}\)
27 tháng 3 2021
E =16+112+120+130+142+156
E=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
E=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{1}-...+\dfrac{1}{7}-\dfrac{1}{8}\)
E=\(\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)
26 tháng 3 2018
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\\ 2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\\ 2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\\ A=1-\dfrac{1}{2^9}=\dfrac{511}{512}\)
\(B=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\\ 3B=\dfrac{3}{4}+\dfrac{3}{12}+\dfrac{3}{36}+\dfrac{3}{108}+\dfrac{3}{324}+\dfrac{3}{972}\\ 3B=\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\\ 3B-B=\left(\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\right)-\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\\ 2B=\dfrac{3}{4}-\dfrac{1}{972}=\dfrac{182}{243}\\ B=\dfrac{364}{243}\)
TV
19 tháng 7 2017
\(D=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=1-\dfrac{1}{6}=\dfrac{5}{6}\)
18 tháng 9 2021
a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
=0
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
KL
3 tháng 8 2017
a, (\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)).10 - x = 0
<=> \(\dfrac{5}{6}.10-x=0\)
<=> \(\dfrac{25}{3}-x=0\)
<=> x = \(\dfrac{25}{3}\) (thỏa mãn)
@Hoàng Mạnh Quân
\(E=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{972}\)
\(\dfrac{1}{3}E=\dfrac{1}{3}\cdot\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{972}\right)\)
\(\dfrac{1}{3}E=\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{90}+...+\dfrac{1}{2916}\)
\(4\cdot\dfrac{1}{3}E=4\cdot\left(\dfrac{1}{12}+\dfrac{1}{36}+...+\dfrac{1}{2916}\right)\)
\(\dfrac{4}{3}E=\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{243}\)
\(\dfrac{4}{3}E=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^5}\)
\(\dfrac{4}{3}E=\dfrac{3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^5}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^5}\right)}{2}\)
\(\dfrac{4}{3}E=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^4}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^5}\right)}{2}\)
\(\dfrac{4}{3}E=\dfrac{1-\dfrac{1}{3^5}}{2}\)
\(\dfrac{4}{3}E=\dfrac{1}{2}\cdot\dfrac{3^5-1}{3^5}\)
\(E=\dfrac{3^5-1}{2\cdot3^5}\cdot\dfrac{3}{4}\)
\(E=\dfrac{3^5-1}{8\cdot3^4}\)