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

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

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

a = 2004 x (4+1+1+1+1)

=2004 x 8=16032

b = 1995 x (1+1+3)

= 1995 x 5 =9975

a, 16032

b, 9975

a,(tự viết đề)

= 2004 x (4+1+1+1+1)

=2004 x 8

=16032

b,(tự viết đề)

= 1995 x (1+1+3)

= 1995 x 5

=9975

a) Ta có: S = 1 - 2 - 3 + 4 + 5 - 6 - 7+  8 + ... + 2001 - 2002 - 2003 + 2004

\(\Rightarrow\)              S = (1 - 2 - 3 + 4) + (5 - 6 - 7+  8) + ... + (2001 - 2002 - 2003 + 2004)

\(\Rightarrow\)             S = (-4 + 4) + (-8 + 8) + ... + (-2004 + 2004)

\(\Rightarrow\)              S = 0 + 0 + ... + 0

\(\Rightarrow\)              S = 0

Câu b): sAI ĐỀ

Giống như trên

=(1-2-3+4)+(5-6-7+8)+...+(2005-2006-2007+2008)+2009

=2009

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

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

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

(1-1/2)x(1-1/3)x(1-1/4)x(1-1/5)x.......x (1-1/2003)x(1-1/2004)

=1/2 x 2/3 x 3/4 x 4/5 x.....x2002/2003 x 2003/2004

=\(\frac{1\times2\times3\times4\times...\times2002\times2003}{2\times3\times4\times5....\times2003\times2004}\)

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

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

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

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

   \(=\frac{1}{2004}\)

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

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

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

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

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