\(\frac{2003\cdot1999-2003\cdot999}{2004\cdot999+1004}\)

\(=\frac{2003\cdot\left(1999-999\right)}{2004\cdot\left(999+1\right)}\)

\(=\frac{2003\cdot1000}{2004\cdot1000}\)

\(=\frac{2003}{2004}\)

\(\dfrac{2003\times1999-2003\times999}{2004\times999+1004}=\) \(\dfrac{2003\times\left(1999-999\right)}{\left(1000+1004\right)\times999+1004}\)

\(=\dfrac{2003\times1000}{1000\times999+1004\times999+1004}\)

\(=\dfrac{2003\times1000}{1000\times999+1004\times\left(999+1\right)}\)

\(=\dfrac{2003\times1000}{1000\times999+1004\times1000}\)

\(=\dfrac{2003\times1000}{1000\times\left(999+1004\right)}\)

\(=\dfrac{2003}{2003}=1\)

2003x1999-2003x999

= 2003x(1999 - 999 )

= 2003 x 1000

= 2003000

2004 x 999 + 1004

= ( 2003 + 1 ) x 999 + 1004

= 2003 x 999 + 999 + 1004

=  2003 x 999 +2003

= 2003 x ( 999 + 1 )

= 2003 x 1000

= 2003000

\(\frac{2003\times1999-2003\times999}{2004\times999+1004}\)

\(=\frac{2003\times\left(1999+999\right)}{2004\times999+1004}\)

\(=\frac{2003\times2998}{2001996+1004}\)

\(=\frac{60004994}{2003000}\)

dãy số hay phân số vậy bạn nếu cả hai thì hai bạn kia đều đúng

=0,9990034933

ket qua = 0,9990034933

Ta có : 

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\frac{50}{51}\)

\(A=\frac{25}{17}\)

Vậy \(A=\frac{25}{17}\)

Chúc bạn học tốt ~ 

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\frac{50}{51}\)

\(A=\frac{25}{17}\)

\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)

\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(B=\frac{21}{4}.33.\frac{4}{21}\)

\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)

\(B=33\)

\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(C=\frac{1}{2}.\frac{98}{99}\)

\(C=\frac{49}{99}\)

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{21}\)

\(A=1-\frac{1}{51}\)

\(A=\frac{51}{51}-\frac{1}{51}\)

\(A=\frac{50}{51}\)

\(A=\frac{21}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(A=\frac{21}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{30.101}+\frac{33.101}{42.101}\right)\)

\(A=\frac{21}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(A=\frac{21}{4}.33\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(A=\frac{21}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(A=\frac{21}{4}.33\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(A=\frac{21}{4}.33\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(A=\frac{21}{4}.33.\frac{4}{21}\)

\(A=33\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(A=\frac{1}{2}.\frac{98}{99}\)

\(A=\frac{49}{99}\)

\(\frac{1\times2\times25}{6\times5\times7}\)\(=\frac{1\times2\times5\times5}{2\times3\times5\times7}=\frac{5}{3\times7}=\frac{5}{21}\)

1/6×2/5×25/7=

rut gon:25(tử)÷5 (mâu), 6 (tử)÷2 (mau)

được

1/3×1/1×5/7=5/21

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)

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

\(=1-\frac{1}{9}=\frac{8}{9}\)

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)

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

\(=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)-\frac{1}{9}\)

\(=1-\frac{1}{9}=\frac{8}{9}\)

 ~ Hok tốt ~

mình vẫn không hiểu

65/18 bạn nhé

bằng 3/11/18 hay là 65/18