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nói chung là :

làm đc vế a là làm đc vế b

$A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$ 

$=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}$ 

$=\frac{31}{32}$ 

Xét: $1-A=1-\frac{31}{32}=\frac{1}{32}$ 

Xét: $1-\frac{2023}{2024}=\frac{1}{2024}$ 

Vì: $32<2024$ 

$\Rightarrow \frac{1}{32}>\frac{1}{2024}$ 

$\Rightarrow 1-\frac{1}{32}<1-\frac{1}{2024}$ 

$\Rightarrow \frac{31}{32}<\frac{2023}{2024}$ 

$Vậy:A<\frac{2023}{2024}$ 

\(A=11+14+17+...+62+65\)

Số số hạng của \(A\)là 

\(\left(65-11\right)\div3+1=19\)(số hạng)

Tổng của \(A\)là: 

\(\left(11+65\right)\times19\div2=722\)

Đáp số: 722

\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(B=\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+.....+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\times\frac{1}{2}\)

\(B=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{9}-\frac{1}{11}\right)\times\frac{1}{2}\)

\(B=\left(1-\frac{1}{11}\right)\times\frac{1}{2}\)

\(B=\frac{10}{11}\times\frac{1}{2}\)

\(B=\frac{5}{11}\)

\(C=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+\frac{3}{154}\)

\(C=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)

\(C=\left(\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+....+\frac{3}{11}-\frac{3}{14}\right)\div3\)

\(C=\left(\frac{3}{2}-\frac{3}{14}\right)\div3\)

\(C=\frac{9}{7}\div3\)

\(C=\frac{3}{7}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(B=1-\frac{1}{11}\)

\(B=\frac{10}{11}\)

\(b=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(b=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(b=1-\frac{1}{11}\)

\(b=\frac{10}{11}\)

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

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

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

(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

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

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

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

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

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

   \(=\frac{1}{2004}\)

Đặt B = 2004+2003/2+2002/3+...+1/2004 B có 2004 phân số tách số 2004 = 1+1+1+...+1(2004 số 1) ghép 2004 số 1 vào từng nhóm như sau: B=(1+ 2003/2)+ (1+ 2002/3)+...+(1+1/2004) +1 B = 2005/2+2005/3+......+2005/2004+2005/2005 B = 2005x(1/2+1/3+....+1/2004+1/2005) Vậy A = 2005

Đặt B = 2004+2003/2+2002/3+...+1/2004
B có 2004 phân số
tách số 2004 = 1+1+1+...+1(2004 số 1)
ghép 2004 số 1 vào từng nhóm như sau:
B=(1+ 2003/2)+ (1+ 2002/3)+...+(1+1/2004) +1
B = 2005/2+2005/3+......+2005/2004+2005/2005 
B = 2005x(1/2+1/3+....+1/2004+1/2005)
Vậy A = 2005

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

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

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

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

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