\(\frac{2003}{1\cdot2}+\frac{2003}{2\cdot3}+...+\frac{2003}{2002\cdot2003}\)

\(=2003\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2002\cdot2003}\right)\)

\(=2003\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{2002}-\frac{1}{2003}\right)\)

\(=2003\cdot\left(1-\frac{1}{2003}\right)\)

\(=2003\cdot\frac{2002}{2003}\)

\(=\frac{2003\cdot2002}{2003}\)

\(=2002\)

714n nha

( 3+5+9+6+85+99+572+25 ). n = 714n

Quãng đường vật đó đi đc sau 1,5 giờ là:

       20 x 1,5 = 30 (km)

                     Đáp số :30 km.

\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(A< 1-\frac{1}{10}=\frac{9}{10}\)

\(=>A>\frac{65}{132}\)

2/3 x ( 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/70.73 - 2x = 1

2/3 x ( 1 - 1/4 + 1/4 - 1/7 + ... + 1/70 - 1/73 ) - 2x = 1

2/3 x ( 1- 1/73 ) - 2x = 1

2/3 x 72/73 - 2x = 1

48/73 - 2x = 1

2x = 48/73 - 1

2x = -25/73

x = -25/73 : 2 

x = -25/146 

Vậy x = -25/146

Tk nha !!