tính
B = \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
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IA
4
17 tháng 6 2017
Thử làm xem , không biết đúng không nhé 
!
\(F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\)
\(F=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(F=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)
\(F=\dfrac{1}{3}.\dfrac{10}{33}\)
\(F=\dfrac{10}{99}\)
17 tháng 6 2017
\(F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+.....+\dfrac{1}{990}\)
\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+......+\dfrac{1}{30.33}\)
\(F=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+.....+\dfrac{1}{30}-\dfrac{1}{33}\)
\(F=\dfrac{1}{3}-\dfrac{1}{33}\)
\(F=\dfrac{10}{33}\)
17 tháng 11 2023
\(\dfrac{1}{x-3}-\dfrac{1}{x}=\dfrac{x-\left(x-3\right)}{x\left(x-3\right)}=\dfrac{x-x+3}{x\left(x-3\right)}=\dfrac{3}{x\left(x-3\right)}\)
\(B=\dfrac{1}{x^2-3x}+\dfrac{1}{x^2-9x+18}+\dfrac{1}{x^2-15x+54}+\dfrac{1}{x^2-21x+108}\)
\(=\dfrac{1}{x\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-6\right)}+\dfrac{1}{\left(x-6\right)\left(x-9\right)}+\dfrac{1}{\left(x-9\right)\left(x-12\right)}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{x\left(x-3\right)}+\dfrac{3}{\left(x-3\right)\left(x-6\right)}+\dfrac{3}{\left(x-6\right)\left(x-9\right)}+\dfrac{3}{\left(x-9\right)\left(x-12\right)}\right)\)
\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-6}-\dfrac{1}{x-6}+\dfrac{1}{x-9}-\dfrac{1}{x-9}+\dfrac{1}{x-12}\right)\)
\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-12}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-\left(x-12\right)+x}{x\left(x-12\right)}\)
\(=\dfrac{4}{x\left(x-12\right)}\)
1 tháng 5 2018
a, A = 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/2017 - 1/2018
A = 1 - 1/2018 = 2017/2018
b, B = 5/2 . ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2016 -1/2018)
B= 5/2 . ( 1/2 - 1/ 2018 )
B = 504/1009
c, C = 1/3.6 + 1/ 6.9 + 1/ 9.12 + ... + 1/ 30.33
C= 1/3 - 1/6 + 1/6 - 1/ 9 + 1/9 - 1/12 + ... + 1/30 - 1/33
C = 1/3 - 1/33
C= 10/33
1 tháng 5 2018
phan B mk quên nhân với 5/2
lấy 5/2 . 504/1009 = 1260/1009
1 tháng 10 2017
\(b,C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\\ =\dfrac{1}{3}-\dfrac{1}{33}\\ =\dfrac{11}{33}-\dfrac{1}{33}=\dfrac{10}{33}\)
1 tháng 10 2017
a.F=\(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{2008.2010}\)
F=\(\dfrac{2.2}{2.4}\)+\(\dfrac{2.2}{4.6}\)+\(\dfrac{2.2}{6.8}\)+...+\(\dfrac{2.2}{2008.2010}\)
F=2.(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{2008.2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{2010}\))
F=\(\dfrac{1004}{1005}\)
9 tháng 4 2017
a) Ta có :
\(A=1+2+2^2+2^3+....................+2^{2010}\) (\(2010\) số hạng)
\(2A=2+2^2+............+2^{2010}+2^{2011}\)
\(\Rightarrow2A-A=\left(2+2^2+..........+2^{2011}\right)-\left(1+2+.............+2^{2010}\right)\)
\(A=2^{2011}-1\)
b) Ta có :
\(B=1-3+3^2-3^3+...............+3^{100}\)(\(100\) số hạng)
\(3B=3-3^2+3^3+.....+3^{99}-3^{100}+3^{101}\)
\(\Rightarrow3B+B=\left(1-3+.......+3^{100}\right)+\left(3-3^2+....-3^{100}+3^{101}\right)\)
\(4B=3^{101}+1\)
~ Chúc bn học tốt ~
Q
9 tháng 4 2017
2)
\(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
\(=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+\dfrac{3}{9.12}+...+\dfrac{3}{30.33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}.\dfrac{10}{33}\)
\(=\dfrac{10}{99}\)
16 tháng 4 2017
Bài 1:
Ta có:
\(A=\dfrac{1}{3}-\dfrac{1}{18}-\dfrac{1}{54}-\dfrac{1}{108}-\dfrac{1}{270}-\dfrac{1}{378}\)
\(=\dfrac{1}{3}-\left(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+\dfrac{1}{270}+\dfrac{1}{378}\right)\)
\(=\dfrac{1}{3}-\left(\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{18.21}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{18}-\dfrac{1}{21}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{21}\right)=\dfrac{1}{3}-\dfrac{1}{3}.\dfrac{6}{21}\)
\(=\dfrac{1}{3}-\dfrac{2}{21}=\dfrac{5}{21}\)
Vậy \(A=\dfrac{5}{21}\)
Bài 2:
Ta có: \(51x+26y=2000\)
Mà \(\left\{{}\begin{matrix}26y⋮2\\2000⋮2\end{matrix}\right.\) \(\Leftrightarrow51x⋮2\)
\(\left(51;2\right)=1\Rightarrow x⋮2\)
Mặt khác \(x\) là số nguyên tố nên \(x=2\)
Khi đó:
\(51.2+26y=2000\Leftrightarrow y=73\) (thỏa mãn)
Vậy các số nguyên tố \(\left(x,y\right)=\left(2;73\right)\)
26 tháng 5 2022
a: \(5\sqrt{2}-8\sqrt{3}+30\sqrt{3}-6\sqrt{3}=5\sqrt{2}+16\sqrt{3}\)
b: \(=14\sqrt{3}-\dfrac{3}{32}\cdot8\sqrt{3}+\dfrac{4}{18}\cdot9\sqrt{3}-\dfrac{1}{10}\cdot10\sqrt{3}\)
\(=14\sqrt{3}-\dfrac{3}{4}\sqrt{3}+2\sqrt{3}-1\sqrt{3}=\dfrac{57}{4}\sqrt{3}\)
c: \(=\dfrac{-1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{22}\cdot11\sqrt{3}+2\sqrt{3}\)
\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{1}{2}\sqrt{3}+2\sqrt{3}=-\dfrac{7}{6}\sqrt{3}\)
d: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)
\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)
BX
1
5 tháng 2 2023
\(2.B=\dfrac{2}{6}+\dfrac{2}{14}+\dfrac{2}{60}+...+\dfrac{2}{990}\)
\(2B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{9.10.11}\)
\(2B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{9.10}-\dfrac{1}{10.11}\)
\(2B=\dfrac{1}{1.2}-\dfrac{1}{10.11}\)
\(B=\dfrac{27}{110}\)
\(B=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\)
\(\Rightarrow B=\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{30.33}\)
\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{30.33}\right)\)
\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)
\(\Rightarrow B=\dfrac{1}{3}.\dfrac{10}{33}\)
\(\Rightarrow B=\dfrac{10}{99}\)
Vậy...
\(B=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
\(\Leftrightarrow B=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+..+\dfrac{1}{30.33}\)
\(\Leftrightarrow B=\left(\dfrac{1}{3}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{9}\right)+...+\left(\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{30}-\dfrac{1}{33}\)
\(\Leftrightarrow B=\dfrac{1}{3}-\dfrac{1}{33}\)
\(\Leftrightarrow B=\dfrac{10}{33}\).