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

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

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

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

\(=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2002}{2003}x\frac{2003}{2004}=\frac{1x2x3x...x2002x2003}{2x3x4x...x2003x2004}=\frac{1}{2004}\)

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

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

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

ta sẽ ra được kết quả qua cách giảm ước của cả tử và mẫu . vậy cuối cùng nhìn lại trên tử còn 1 mẫu thì còn 2004 vậy phân số ra được là 1/2004

Tính nhanh : 

( 1 - 1/2 ) × ( 1 - 1/3 ) × ( 1 - 1/4 ) × ... × ( 1 - 1/2003 ) × ( 1 - 1/2004 ) 

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

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

= 1/2004

(1 - 1/ 2) x (1 - 1/3) x ( 1 - 1/4 ) x ... x (1 - 1/2003) × ( 1 - 1/ 2004 )

                     Giải

( 1 - 1/2 ) × ( 1 - 1/3 ) × ( 1 - 1/4 ) × ... × ( 1 - 1/2003 ) × ( 1 - 1/2004 ) 

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

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

= 1/2004

\(B=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2002}{2003}x\frac{2003}{2004}\)

\(B=\frac{1x2x3x4x...x2002x2003}{2x3x4x5x...x2003x2004}\)

Rút gọn các thừa số ở tử và mẫu ta được:

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

                                                                              Đ/S:\(\frac{1}{2004}\)

Ta có:

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

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

\(=\frac{1.2....2002.2003}{2.3....2003.2004}\)

Đơn giản hết sẽ là:

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

B = \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x......x\left(1-\frac{1}{2003}\right)x\left(1-\frac{1}{2004}\right)\)

B = \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x.........x\frac{2002}{2003}x\frac{2003}{2004}\)

B = \(\frac{1x2x3x4x..........x2002x2003}{2x3x4x5x.....x2003x2004}\)

Sau khi trực tiêu các số thì ta có:

B = \(\frac{1}{2004}\)

làm ơn tách ra giùm mk

????

0

Đề bài

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

= \(1\times2\times3\times4\times...\times2002\times2003/2\times3\times4\times5\times...2003\times2004\)

= \(\frac{1}{2004}\)

Đề bài 

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

=   1/2004

K nha 

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

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

1/2004

\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2004}\right)\)

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

\(\Rightarrow A=\frac{1\cdot2\cdot3...2003}{2\cdot3\cdot4...2004}=\frac{1}{2004}\)