\(\frac{1}{3}+\frac{2}{15}+\frac{5}{35}+\frac{4}{63}+\frac{5}{99}\)

\(=\frac{1155}{3465}+\frac{462}{3465}+\frac{495}{3465}+\frac{220}{3465}+\frac{175}{3465}\)

\(=\frac{2507}{3465}\)

\(X-\left(\frac{31}{5}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\frac{9}{13}\)

\(X-\left(\frac{31}{5}+\frac{31}{3\cdot5}+\frac{31}{5\cdot7}+\frac{31}{7\cdot9}+\frac{31}{9\cdot11}+\frac{31}{11\cdot13}\right)=\frac{9}{13}\)

\(X-\left[\frac{31}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)+\frac{31}{5}\right]=\frac{9}{13}\)

\(X-\left[\frac{31}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)+\frac{31}{5}\right]=\frac{9}{13}\)

\(X-\left[\frac{31}{2}\cdot\frac{10}{39}+\frac{31}{5}\right]=\frac{9}{13}\)

\(X-\frac{1984}{195}=\frac{9}{13}\)

\(\Rightarrow X=\frac{9}{13}+\frac{1984}{195}=\frac{163}{15}\)

163 / 15

mk nghĩ v

Toán lớp 3?Sao giống lớp 4,5 hơn!

Ta có:

\(A=\frac{5}{15}+...+\frac{5}{399}\)

\(\Rightarrow A=5.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)

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

\(\Rightarrow A=\frac{5}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(\Rightarrow A=\frac{5}{7}\)

\(A=\frac{5}{15}+\frac{5}{35}+\frac{5}{63}+...+\frac{5}{399}\)

\(A=\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+\frac{5}{7\cdot9}+...+\frac{5}{19\cdot21}\)

\(A=\frac{5}{3\cdot5\cdot2}+\frac{5}{5\cdot7\cdot2}+\frac{5}{7\cdot9\cdot2}+...+\frac{5}{19\cdot21\cdot2}\)

\(A=\frac{5}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{19\cdot21}\right)\)

\(A=\frac{5}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(A=\frac{5}{2}\cdot\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(A=\frac{5}{2}\cdot\left(\frac{7}{21}-\frac{1}{21}\right)\)

\(A=\frac{5}{2}\cdot\frac{6}{21}\)

\(A=\frac{30}{42}=\frac{5}{7}\)

Đăt S=1/15+1/35+1/63+1/99+...+1/2915+1/3135

         =1/3.5+1/5.7+1/7.9+1/9.11+...+1/53.55+1/55.57

         =1/2(2/3.5+2/5.7+2/7.9+...+2/53.55+2/55.57)

         =1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)

         =1/2(1/3-1/57)

         =1/2(19/57-1/57)

         =1/2.18/57

         =3/19

Vậy 1/15+1/35+1/63+1/99+...+1/2915+1/3135=3/19

Mik viết thế này mong bạn thông cảm nha!!

chúc bạn hok tốt!!

Bạn nhớ k cho mik một cái đúng nha!!

Đặt \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)

\(\Leftrightarrow A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+....+\frac{1}{53\cdot55}+\frac{1}{55\cdot57}\)

\(\Leftrightarrow2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{53\cdot55}+\frac{2}{55\cdot57}\)

\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)

\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)

\(\Leftrightarrow A=\frac{6}{19}:2=\frac{3}{19}\)

1/15 + 1/35 + 1/63 + 1/99 + ...+ 1/2915 + 1/3135

= 1/2.(2/15 + 2/35 + 2/63 + 2/99 + ... + 2/2915 + 2/3135)

=1/2.( 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + ... + 2/53.55 + 2/55.57)

=1/2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)

=1/2.(1/3-1/57)

=1/2.6/19

=3/19

2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

      = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)

      =\(1-\frac{1}{15}=\frac{14}{15}\)

\(\Rightarrow S=\frac{7}{15}\)

a. Ta có:A= 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15

A=1/2(1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)

A=1/2(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15)

A=2(1-1/15)

A=1/2.14/15

A=7/15

phần b nè

pt \(\Rightarrow90-6ab=3a\)\(\Leftrightarrow3a\left(b+2\right)=90\)vì b>0 \(\Leftrightarrow a=\frac{30}{b+2}\)mà a,b \(\inℕ^∗\)

\(\Rightarrow\)b+2\(\inƯ\left(30\right)\)MÀb\(\inℕ^∗\)\(b+2\in\left\{3;5;6;10;15;30\right\}\)khi đó tìm đc b \(\rightarrow\)thau vào tìm a . nhớ thử lại vào pt ban đầu nhé 

k cho mk nha mn ^.^

a) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=\frac{1}{2}.\frac{10}{39}\)

\(=\frac{5}{39}\)

a)1/3.5+1/5.7+...+1/11.13

=1/2x(1/3-1/5+1/5-1/7+...+1/11-1/13)

=1/2x(1/3-1/13)

=1/2x10/39

=5/39

quá dễ cái này lớp 4 mik hok rùi thật

giúp mình ik nhá

\(A=5\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\right)\)

\(=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\right)\)

\(=5\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{81-80}{80.81}\right)\)

\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\right)\)

\(=5\left(1-\frac{1}{81}\right)=\frac{5.80}{81}=\frac{400}{81}\)

b)

\(B=7\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{483}\right)\)

\(=7.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{21.23}\right)\)

=> \(2.B=7\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{21.23}\right)\)

\(=7\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{23-21}{21.23}\right)\)

\(=7.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{21}-\frac{1}{23}\right)\)

\(=7\left(\frac{1}{3}-\frac{1}{23}\right)=\frac{7.20}{69}=\frac{140}{69}\)

=> \(B=\frac{140}{69}:2=\frac{70}{69}\)

a) SỬA LẠI ĐỀ : A = \(\frac{5}{2}+\frac{5}{6}+\frac{5}{12}+...+\frac{5}{6480}\)

       = \(5.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\right)\)

       =\(5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\right)\)

        = \(5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\right)\)

        = \(5.\left(1-\frac{1}{81}\right)\)

        = \(5.\frac{80}{81}\)

        = \(\frac{400}{81}\)

b) B = \(\frac{7}{15}+\frac{7}{35}+\frac{7}{63}+...+\frac{7}{483}\)

        = \(\frac{7}{3.5}+\frac{7}{5.7}+\frac{7}{7.9}+...+\frac{7}{21.23}\)

        = \(\frac{7}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{21.23}\right)\)

        = \(\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{23}\right)\)

        = \(\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{23}\right)\)

        = \(\frac{7}{2}.\frac{20}{69}\)

        = \(\frac{70}{69}\)

a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) 

vì  ( 125125 x 127 - 127127 x 125 ) =[125125 x (125+2)] - 127127 x 125 ) =>125125 x (125+2)=125.125125+125125.2=125125.125+250250=125125.125+125.2002=125.(125125+2002)=125.127127

=> ( 125125 x 127 - 127127 x 125 )=127127.125-127127.125=0

=>  (1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) =0

a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) 

= ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 )  x 0

= 0

b, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)

= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)

= \(\frac{1}{3}\)- \(\frac{1}{15}\)

= \(\frac{4}{15}\)

1)

a)

\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)

\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)

\(\frac{-20}{5}< x< \frac{-3}{10}\)

\(\frac{-40}{10}< x< \frac{-3}{10}\)

\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)

\(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)

\(\frac{25}{3}< x< \frac{-4}{7}.\frac{1}{1}\)

\(\frac{-25}{3}< x< \frac{-4}{7}\)

\(\frac{-175}{21}< x< \frac{-12}{21}\)

\(\Rightarrow Z\in\left\{-13;-14;-15;-16;...;-174\right\}\)

2)

a)

\(\left(\frac{9}{10}-\frac{15}{16}\right).\left(\frac{5}{12}-\frac{11}{15}-\frac{7}{20}\right)\)

\(=\left(\frac{72}{80}-\frac{75}{80}\right).\left(\frac{25}{60}-\frac{44}{60}-\frac{21}{60}\right)\)

\(=\frac{-3}{80}.\frac{-40}{60}\)

\(=\frac{-1}{-2}.\frac{-1}{-20}\)

\(=\frac{1}{40}\)