\(B=\dfrac{1}{2}x\dfrac{2}{3}x\dfrac{3}{4}x...x\dfrac{2003}{2004}\)

\(B=\dfrac{1}{2004}\)

\(B=\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times....\left(1-\dfrac{1}{2003}\right)\times\left(1-\dfrac{1}{2004}\right)\)

\(B=\dfrac{1}{2}\times\dfrac{2}{3}\times....\times\dfrac{2002}{2003}\times\dfrac{2003}{2004}\) 

\(B=\dfrac{1}{2004}\)

a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)

\(=\dfrac{1}{2004}\)

b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)

\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{10}{3}\)

\(=\dfrac{3}{5}\)

ai trả lời đúng thì mình sẽ tick cho

\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right).\)

\(\Rightarrow A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}....\dfrac{2002}{2003}.\dfrac{2003}{2004}\)

\(\Rightarrow A=\dfrac{1}{2004}\)

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2004}\right)\\ =\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}...\dfrac{2003}{2004}\\ =\dfrac{1}{2004}\)

\(=\frac{1}{2}\frac{2}{3}\frac{3}{4}\frac{4}{5}\frac{5}{6}\frac{6}{7}\frac{7}{8}\frac{8}{9}....\frac{2002}{2003}\frac{2003}{2004}\)

\(=\frac{1}{2004}\)

A=1/2 x 2/3 x 3/4 x ... x 2002/2003 x 2003/2004

=1/2004

XEM LẠI GIÙM MÌNH NHA!

bo ko bt

Giúp mình nhanh nha! Mình sẽ thick người đó

a) x+(x+1)+(x+2)+(x+3)+...+2003=2003

    x+(x+1)+(x+2)+(x+3)+...+2003=2003

     X+(x+1)+(x+2)+(x+3)+...+2002=0

(    Vì ta thấy đây là tổng của một dãy số các số hạng liên tiếp nên day tren co so cuoi la 2002 va tong tat ca bang 0 vi 2003-2003=0 ma)

Goi so so hang cua day so tren la n(nkhac 0)

Suy ra ta co ((2002+x).n):2=0

                    suy ra (2002+x).n=0

                      Mà n khác 0

                       Suy ra 2002+x=0

                                          x=0-2002

                                             x=-2002

                                Vay x=-2002 

        Cậu b bạn làm tương tự nhé!

Neu to co lam sai thi ban thong cam nhe!

a)\(\dfrac{1}{10000}+\dfrac{13}{10000}+\dfrac{25}{10000}+...+\dfrac{97}{10000}+\dfrac{109}{10000}\)

\(=\dfrac{1+13+25+...+97+109}{10000}\)

\(=\dfrac{\left(1+109\right)\left[109-1\right]:12+1}{20000}\)

\(=\dfrac{110.10}{20000}=\dfrac{11}{200}\)

b)\(\dfrac{4}{3}\times2019\times0,75\)

=\(\dfrac{4}{3}\times\dfrac{3}{4}\times2019\)

\(=2019\)

c)\(4\times5\times0,25\times\dfrac{1}{5}\times\dfrac{1}{2}\times2\)

\(=\left(4\times\dfrac{1}{4}\right)\left(5\times\dfrac{1}{5}\right)\left(2\times\dfrac{1}{2}\right)\)

\(=1\times1\times1=1\)

Ý d) đặt tính kiểu gì thế ?

2023 rồi còn ai ở đây ko 😢😢😢

B= 1/2 x 2/3 x 3/4 x ...........x 2002/2003 x 2003/2004

1 x 2 x 3 x 4 x .............x 2002 x 2003

2 x 3 x 4 x .............x 2003 x 2004

1

 2004