Câu 1: Tìm tổng C:
C = ( a + 2 x b ) + ( b + 2 x c ) + ( c + 2 x a )
với a + b + c = 15.
Câu 2: Tìm tổng D:
D = \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
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<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)
<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=> \(\frac{2}{15}+x=\frac{17}{15}\)
=> 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
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
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}\cdot\frac{14}{15}\)
\(=\frac{7}{15}\)
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
a) \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}\)
= \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\)
= \(\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
= \(\frac{1}{2}.\frac{4}{15}\)
= \(\frac{2}{15}\)
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
Bài 1:
a)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
b)\(=1008\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=1008\cdot\left(1-\frac{1}{2017}\right)\)
\(=1008\cdot\frac{2016}{2017}\)\(=\frac{290304}{31}\)Bài 2:
a)\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}\)
\(=\frac{2}{7}\)
b)\(B=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{25.28}\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\cdot\frac{6}{28}\)
\(=\frac{15}{14}\)
Bài 3:
a)Đặt \(A=-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}\)
\(=-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)\right]\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{55}\right)\right]\)
\(=-\left[10\cdot\frac{4}{55}\right]\)
\(=-\frac{8}{11}\).Thay vào ta có: \(x-\frac{8}{11}=\frac{2}{9}\)
\(\Leftrightarrow x=\frac{94}{99}\)
b)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(x+1=18\)
\(x=17\)
a)Ta có:
A= 1/15+1/35+1/63+1/99+1/143
A= 1/3.5+1/5.7+1/7.9+1/9.11+1/11.13
2A= 2/3.5+2/5.7+2/7.9+2/9.11+2/11.13
2A= 1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
Đơn giản đi ta được:
2A= 1/3-1/13
2A= 10/39
A= 5/39
Vậy A= 5/39
b) Để A và B có giá trị bằng nhau thì:
\(\frac{3}{4}\cdot x+7=\frac{4}{3}\cdot x-35\)
\(7+35=\frac{4}{3}\cdot x-\frac{3}{4}\cdot x\)
\(42=\frac{7}{12}\cdot x\)
\(x=42:\frac{7}{12}\)
\(x=72\)
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, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)
= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)
= \(\frac{1}{3}\)- \(\frac{1}{15}\)
= \(\frac{4}{15}\)
câu 2:
= 6/13
Các bạn nêu rõ cách làm từng bài giúp mình nhé! Thanks ^-^!