\(\frac{7}{15}\)nha ban

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mik nha

Dấu \(.\)là dấu nhân 

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

\(=\frac{1}{2}.\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}+\frac{2}{195}\right)\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\frac{14}{15}\)

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

~ Ủng hộ nhé 

Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

Suy ra ; \(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

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

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

=> A = \(\frac{14}{15}:2=\frac{14}{15}.\frac{1}{2}=\frac{7}{15}\)

Gọi dãy trên là A

\(\Leftrightarrow A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)

\(\Leftrightarrow2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\)

\(\Leftrightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)

\(\Leftrightarrow2A=1-\frac{1}{15}\)

\(\Leftrightarrow2A=\frac{14}{15}\)

\(\Leftrightarrow A=\frac{7}{15}\)

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

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\frac{14}{15}\)

\(=\frac{14}{30}=\frac{7}{15}\)

a)

\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

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

\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(=2\left(1-\frac{1}{15}\right)\)

\(=2.\frac{14}{15}\)

\(=\frac{28}{15}\)

b)

\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)

\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}+\frac{2}{10.11}+\frac{2}{11.12}\)

                                                                                         \(...\)

câu a mình nhầm phải là ½.(…)

<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)

<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)

<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)

<=> \(\frac{2}{15}+x=\frac{17}{15}\)

=> x = 1

(1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)+x=17/15

[2.(1/3-1/5+1/5-1/7+...+1/13-1/15)]+x=17/15

[2.(1/3-1/15)]+x=17/15

(2.4/15)+x=17/15

6/15+x=17/15

x=17/15-6/15

x=11/15

a=8-\(\frac{8}{3.5}-\frac{8}{5.7}-\frac{8}{7.9}-\frac{8}{9.11}-\frac{8}{11.13}-\frac{8}{13.15}\)

a=8-\(\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}-\frac{1}{13}+\frac{1}{15}\)

a=8-1/3+1/15=126/15

= 13/15 nha bn

duyệt đi

A=1-1/15-1/35-1/63-1/99-1/143-1/195

=1-1/3.5-1/5.7-1/7.9-1/9.11-1/11.13-1/13.15

=1-1/2(2/3.5-2/5.7-2/7.9-2/9.11-2/11.13-2/13.15)

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

=1-1/2.(1/3-1/15)

=1-1/2.4/15

=1-2/15=13/15

A = 1-1/3*5-1/5*7-...-1/13*15

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

1/2 A = 1-1/3-1/15

  1/2   A = 2/3-1/15

1/2 A = 3/5

a = 3/5 * 2= 6/5

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 ^.^

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

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)

\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{15}\right)\)

\(=\frac{1}{2}\cdot\frac{14}{15}\)

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

Sửa đề chút nhé:

\(\left(1+3+5+7+...+2009+2011\right).\left(125125.127-127127.125\right)\)

\(=\left(1+3+5+7+...+2009+2011\right).\left(125.1001.127-127.1001.125\right)\)

\(=\left(1+3+5+7+...+2009+2011\right).0\)

\(=0\)

Ý b tham khảo bài bạn nguyen thi thuy linh nhé

\(\text{Tính nhanh : }\)

\(a,\text{ }1+3+5+7+9+\text{...}+2007+2009+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=\left\{\left(2009-1\right)\text{ : }2+1\right\}\cdot\left(2009+1\right)\text{ : }2+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=1005\cdot2010\text{ : }2+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=2020050\text{ : }2+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=1010025+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=1010025+2011\cdot\left(15890875+15890875\right)\)

\(=1010025+2011\cdot15890875\cdot2\)

\(=1010025+31956549625\cdot2\)

\(=1010025+63913099250\)

\(=63914109275\)

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

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{15}\)

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

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

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

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

= \(\frac{1}{3}\)- \(\frac{1}{15}\)

= \(\frac{4}{15}\)