1) =1/2 x 2/3 x 3/4 x 4/5 x .... x 2002/2003 x 2003/2004

=1/2004

2) 1/2 x X-3/4=5/6

1/2 x X =3/4+5/6

1/2 x X =19/12

X=19/6

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)

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

\(=\frac{1.2.3...2002.2003}{2.3.4...2003.2004}=\frac{1}{2004}\)

\(\frac{1}{2}.x-\frac{3}{4}=\frac{5}{6}\)

\(\frac{1}{2}.x=\frac{5}{6}+\frac{3}{4}\)

\(\frac{1}{2}.x=\frac{10}{12}+\frac{9}{12}=\frac{19}{12}\)

\(x=\frac{19}{12}:\frac{1}{2}\)

\(x=\frac{19}{12}.2=\frac{19}{6}\)

(1-1/2) x (1-1/3) x (1-1/4) x...x (1-1/2004)

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

1/2 x X -3/4 = 5/6

=> 1/2 x X = 5/6 +3/4 = 19/12

=> x= 19/12 : 1/2

=> x=19/6

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

Giúp mình nhanh nha! Mình sẽ thick người đó

\(A=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2002}{2003}x\frac{2003}{2004}\)

\(A=\frac{1x\left(2x3x4x...x2002x2003\right)}{\left(2x3x4x...x2002x2003\right)x2004}=\frac{1}{2004}\)

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!

bn ơi 1/3 ko đc bn ơi

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

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

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

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