\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}+\frac{1}{21.23}\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}+\frac{2}{21.23}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}+\frac{1}{21}-\frac{1}{23}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{23}\right)\)

\(A=\frac{1}{2}.\frac{22}{23}\)

\(A=\frac{11}{23}\)

\(\frac{1}{1.3}\)+\(\frac{1}{3\cdot5}\)+................+\(\frac{1}{19.21}\)

Đặt A = \(\frac{1}{1.3}\)+\(\frac{1}{3\cdot5}\)+.............+\(\frac{1}{19.21}\)

Nhân cả 2 vế của A với 2 ta có :

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

2A = \(\frac{1}{1}\)-\(\frac{1}{3}\)+ \(\frac{1}{3}\) -  \(\frac{1}{5}\)+ ............ + \(\frac{1}{19}\)- \(\frac{1}{21}\)

2A = \(\frac{1}{1}\)- \(\frac{1}{21}\)

2A = \(\frac{20}{21}\)

  A = \(\frac{20}{21}\): 2 

  A = \(\frac{10}{21}\)

=1/1.3.5+1/3/5/7+1/5.7.9+......+1/17/19/21

=1/4.(5-1/1.3.5+7-3/3.5.7+.....+21-17/17/19/21

=1/4.(5/1.3.5-1/1.3.5+7/3.5.7-3/3.5.7+.....+21/17.19.21-17/17.19.21

=1/4.(1/1.3-1/3.5+1/3.5-1/5.7+.....+1/17.19-1/19.21)

=1/4.(1/3.1/21.17)

=1/4.3200/9603

= 800/9603

            Chúc bạn học tốt^^

Đặt \(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{17.19.21}\)

\(\Rightarrow4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{17.19.21}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{17.19}-\frac{1}{19.21}\)

\(=\frac{1}{1.3}-\frac{1}{19.21}=\frac{44}{133}\)

\(\Rightarrow A=\frac{44}{133}\div4=\frac{11}{133}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\)

\(=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{21}\right)=\frac{1}{2}.\frac{7-1}{21}=\frac{1}{7}\)

\(\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{19\times21}\)

\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{19\times21}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(=\frac{1}{2}\times\frac{2}{7}\)

\(=\frac{1}{7}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}+\frac{1}{21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)=\frac{1}{2}.\frac{6}{21}=\frac{1}{7}\)

3 x 15 + 21 x 15 + 85 x 5

= 45 + 315 + 425

= 785

15 - 30 + 40 

= 25

21 + 19 - 50 + 10

= 0

\(\dfrac{1}{5}-\dfrac{1}{4}+2\)

\(=-\dfrac{1}{20}+2\)

\(=\dfrac{39}{20}\)

\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)

\(=\dfrac{5}{12}\times\dfrac{1}{4}\)

\(=\dfrac{5}{12}\times\dfrac{3}{12}\)

\(=\dfrac{5}{48}\)

\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)

\(=-\dfrac{9}{20}\)

\(3\times15+21\times15+85\times5\\ =15\times\left(3+21\right)+425\\ =15\times24+425\\ =360+425\\ =785\)

\(15-30+40\\ =\left(15+40\right)-30\\ =55-30\\ =25\)

\(21+19-50+10\\ =\left(21+19\right)-\left(50-10\right)\\ =40-40\\ =0\)

\(\dfrac{1}{5}-\dfrac{1}{4}+2\)

\(=\dfrac{4}{20}-\dfrac{5}{20}+\dfrac{40}{20}\)

\(=\dfrac{\left(4+40\right)}{20}-\dfrac{5}{20}\)

\(=\dfrac{44}{20}-\dfrac{5}{20}\)

\(=\dfrac{39}{20}\)

\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)

\(=\dfrac{5}{12}\times\dfrac{1}{4}\)

\(=\dfrac{5}{48}\)

\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)

\(=\dfrac{2}{20}+\dfrac{4}{20}-\dfrac{15}{20}\)

\(=\dfrac{6}{20}-\dfrac{15}{20}\)

\(=-\dfrac{9}{20}\)

Help meeeeeeeeeee

ai trả lời hộ mình đi

ta có: 1/61>1/72>1/83>1/94=>1/16+1/72+1/83+1/94<1/61*4=1/15

1/16>1/19>1/21=>1/16+1/21+1/1/19<3*1/16<3*1/15=3/15

1/3=5/15

ta được: 1/3+1/16+1/19+1/21+1/61+1/72+1/83+1/94<4/15+3/15+5/15=9/15=3/5 (ĐPCM)

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