\(F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+....+\dfrac{1}{30.33}\)

\(F=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\)

\(F=\dfrac{1}{3}-\dfrac{1}{30}\)

\(F=\dfrac{3}{10}\)

Thử làm xem , không biết đúng không nhé !

\(F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\)

\(F=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(F=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)

\(F=\dfrac{1}{3}.\dfrac{10}{33}\)

\(F=\dfrac{10}{99}\)

\(F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+.....+\dfrac{1}{990}\)

\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+......+\dfrac{1}{30.33}\)

\(F=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+.....+\dfrac{1}{30}-\dfrac{1}{33}\)

\(F=\dfrac{1}{3}-\dfrac{1}{33}\)

\(F=\dfrac{10}{33}\)

\(\dfrac{1}{x-3}-\dfrac{1}{x}=\dfrac{x-\left(x-3\right)}{x\left(x-3\right)}=\dfrac{x-x+3}{x\left(x-3\right)}=\dfrac{3}{x\left(x-3\right)}\)

\(B=\dfrac{1}{x^2-3x}+\dfrac{1}{x^2-9x+18}+\dfrac{1}{x^2-15x+54}+\dfrac{1}{x^2-21x+108}\)

\(=\dfrac{1}{x\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-6\right)}+\dfrac{1}{\left(x-6\right)\left(x-9\right)}+\dfrac{1}{\left(x-9\right)\left(x-12\right)}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{x\left(x-3\right)}+\dfrac{3}{\left(x-3\right)\left(x-6\right)}+\dfrac{3}{\left(x-6\right)\left(x-9\right)}+\dfrac{3}{\left(x-9\right)\left(x-12\right)}\right)\)

\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-6}-\dfrac{1}{x-6}+\dfrac{1}{x-9}-\dfrac{1}{x-9}+\dfrac{1}{x-12}\right)\)

\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-12}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-\left(x-12\right)+x}{x\left(x-12\right)}\)

\(=\dfrac{4}{x\left(x-12\right)}\)

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

phan B mk quên nhân với 5/2

lấy 5/2 . 504/1009 = 1260/1009

a) \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2017.2018}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)

\(=1-\dfrac{1}{2018}=\dfrac{2017}{2018}.\)

b) \(B=\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{2016.2018}\)

\(=2,5\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2016.2018}\right)\)

\(=2,5\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2016}-\dfrac{1}{2018}\right)\)

\(=2,5\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=\dfrac{1260}{1009}.\)

c) \(C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

=>\(3C=\dfrac{3}{3.6}+\dfrac{3}{6.9}+\dfrac{3}{9.12}+...+\dfrac{1}{30.33}\)

\(=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\)

\(=\dfrac{1}{3}-\dfrac{1}{33}=\dfrac{10}{33}\)

=> \(C=\dfrac{10}{33}:3=\dfrac{10}{99}.\)

\(b,C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\\ =\dfrac{1}{3}-\dfrac{1}{33}\\ =\dfrac{11}{33}-\dfrac{1}{33}=\dfrac{10}{33}\)

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=\(\dfrac{1004}{1005}\)

b, C=\(\dfrac{1}{18}\)+\(\dfrac{1}{54}\)+....+\(\dfrac{1}{990}\)

\(\Rightarrow\)C=\(\dfrac{1}{3.6}\)+\(\dfrac{1}{6.9}\)+...+\(\dfrac{1}{30.33}\)

=>3C=3( \(\dfrac{1}{3.6}\)+\(\dfrac{1}{6.9}\)+...+\(\dfrac{1}{30.33}\))

=>3C=\(\dfrac{3}{3.6}\)+\(\dfrac{3}{6.9}\)+....+\(\dfrac{3}{30.33}\)

=> 3C=\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{9}\)+...+\(\dfrac{1}{30}\)-\(\dfrac{1}{33}\)

=> 3C= \(\dfrac{1}{3}\)-\(\dfrac{1}{33}\)

=>3C=\(\dfrac{10}{33}\)

=> C=\(\dfrac{10}{33}\):3

=> C=\(\dfrac{10}{99}\)

a, F=4/2.4+4/4/6+4/6.8+......+4/2008.2010

=> F= 4/2.(2/2.4+2/4.6+2/6.8+......+2/2008/2010

=> F= 4/2. ( 1/2-1/4+1/4-1/6+1/6-1/8+......+1/2008-1/2010

=> F=4/2.( 1/2-1/2010)

=> F= 4/2. 502/1005

=> F= \(\dfrac{1004}{1005}\)

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 ~

2)

\(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

\(=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+\dfrac{3}{9.12}+...+\dfrac{3}{30.33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}.\dfrac{10}{33}\)

\(=\dfrac{10}{99}\)

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 \(\left(x,y\right)=\left(2;73\right)\)

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: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)

\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)

đó bạn

\(2.B=\dfrac{2}{6}+\dfrac{2}{14}+\dfrac{2}{60}+...+\dfrac{2}{990}\)

\(2B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{9.10.11}\)

\(2B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{9.10}-\dfrac{1}{10.11}\)

\(2B=\dfrac{1}{1.2}-\dfrac{1}{10.11}\)

\(B=\dfrac{27}{110}\)