\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\)

\(A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)

\(A=1-\frac{1}{21}\)

\(A=\frac{20}{21}\)

\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}\)

\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+...+\left(1-\frac{1}{9999}\right)\)

\(=\left(1-\frac{1}{1\cdot3}\right)+\left(1-\frac{1}{3\cdot5}\right)+\left(1-\frac{1}{5\cdot7}\right)+\left(1-\frac{1}{7\cdot9}\right)+...+\left(1-\frac{1}{99\cdot101}\right)\)

\(=\left(1+1+1+1+...+1\right)-\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

Có tất cả : (101 - 3) : 2 + 1 = 50 chữ số 1 => (1 + 1 + 1 + .... + 1) = 1 x 50 = 50 

\(\Rightarrow50-\frac{1}{2}\cdot\left(1-\frac{1}{101}\right)\)

\(=50-\frac{1}{2}\cdot\frac{100}{101}=50-\frac{100}{101}=\frac{4950}{101}\)

Vậy \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}=\frac{4950}{101}\)

\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right).\frac{-4}{15}}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)

\(=\frac{\frac{1}{7}\left(\frac{1}{2}-\sqrt{2}+\frac{3\sqrt{2}}{5}\right).\frac{-4}{15}}{\frac{1}{5}\left(\frac{1}{2}+\frac{3\sqrt{2}}{5}-\sqrt{2}\right).\frac{5}{7}}\)

\(=\frac{\frac{1}{7}.\frac{-4}{15}}{\frac{1}{5}.\frac{5}{7}}=\frac{\frac{-4}{105}}{\frac{1}{7}}=\frac{-4}{15}\)

\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\left(\frac{1}{10}+\frac{3.\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)

\(=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\frac{1}{10}.\frac{5}{7}+\frac{3.\sqrt{2}}{25}.\frac{5}{7}-\frac{\sqrt{2}}{5}.\frac{5}{7}}\)

\(=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\frac{1}{14}+\frac{3.\sqrt{2}}{35}-\frac{\sqrt{2}}{7}}\)

\(=\frac{-4}{15}\)

Học tốt

\(a,=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)

\(=1+-1\)

\(=0\)

\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+...+\frac{2}{899}\)
\(=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{29\cdot31}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{31}\)
\(=\frac{1}{3}-\frac{1}{31}\)
\(=\frac{28}{93}\)

\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{899}\)

= \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{29.31}\)

= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{31}\)

= \(\frac{1}{3}-\frac{1}{31}+0+0+...+0\)

= \(\frac{29}{93}\)

C = 1/3 + -3/4 + 3/5 + 1/57 + -1/36 + 1/15 + -2/9

C = ( 1/3 + 1/57 ) + ( -3/4 + -1/36 ) + ( 3/5 + 1/15 ) + -2/9 

C = ( 19/57 + 1/57 ) + ( -27/36 + -1/36 ) + ( 9/15 + 1/15 ) + -2/9 

C = 20/57 + -28/36 + 10/15 + -2/9 

C = 20/57 + -7/9 + 2/3 + -2/9

C = ( 20/57 + 2/3 ) + ( -7/9 + -2/9 )

C = 58/57 + -1 

C = 1/57

D = 1/2 + -1/5 + -5/7 + 1/6 + -3/35 + 1/3 + 1/41

D = ( 1/2 + 1/3 + 1/6 ) + ( -1/5 + -5/7 +-3/35 ) + 1/41

D = ( 3/6 + 2/6 + 1/6 ) + ( -7/35 + -25/35 + -3/35 ) + 1/41

D = 1 + -1 + 1/41

D = 1/41

E = -1/2 + 3/5 + -1/9 + 1/127 + -7/18 + 4/35 + 2/7 

E = ( -1/2 + -1/9 + -7/18 ) + ( 3/5 + 4/35 ) + 1/127 + 2/7

E = ( -9/18 + -2/18 + -7/18 ) + ( 21/35 + 4/35 ) + 1/127 + 2/7

E = -1 + 5/7 + 1/257 + 2/7 

E = -1 + ( 5/7 + 2/7 ) + 1/127

E = -1 + 1 + 1/127

E = 1/127

\(C=\frac{1}{3}+\frac{-3}{4}+\frac{3}{5}+\frac{1}{57}+\frac{-1}{36}+\frac{1}{15}+\frac{-2}{9}.\)

\(C=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{1}{36}+\frac{2}{9}\right)+\frac{1}{57}\)

\(C=1-1+\frac{1}{57}\)

\(C=\frac{1}{57}\)

D=1/2+-1/5+-5/7+1/6+-3/35+1/3+1/41

D=(1/2+1/6+1/3)-(1/5+3/35+5/7)+1/41

D=1-1+1/41

D=1/41

E=-1/2+3/5+-1/9+1/127+-7/18+4/35+2/7

E=(3/5+4/35+2/7)-(1/2+1/9+7/18)+1/127

E=1-1+1/127

E=1/127