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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
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 = \(\frac{14}{15}:2=\frac{14}{15}.\frac{1}{2}=\frac{7}{15}\)
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:
\(\frac{3}{4}\cdot x+7=\frac{4}{3}\cdot x-35\)
\(7+35=\frac{4}{3}\cdot x-\frac{3}{4}\cdot x\)
\(42=\frac{7}{12}\cdot x\)
\(x=42:\frac{7}{12}\)
\(x=72\)
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{7}{15}\)
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{15}\right)=\frac{1}{2}.\frac{14}{15}\)\(=\frac{7}{15}\)
b)\(\frac{1414+1515+...+1919}{2020+2121+...+2525}\)
\(\Rightarrow\frac{101\left(14+15+16+17+18+19\right)}{101\left(20+21+22+23+24+25\right)}\)
\(=\frac{14+15+16+17+18+19}{20+21+22+23+24+25}\)
\(=\frac{\left(19+14\right).6:2}{\left(25+20\right).6:2}=\frac{19+14}{25+20}=\frac{33}{45}=\frac{11}{15}\)
Các bạn nêu rõ cách làm từng bài giúp mình nhé! Thanks ^-^!
a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\cdot\frac{8}{33}\)
\(=\frac{52}{33}\)
a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99
A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)
A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)
A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
A= 13/2 ( 1/3 - 1/11)
A= 13/2 . 8/33
A= 52/33
\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)
\(\Leftrightarrow2x-4,36=1\)
\(\Leftrightarrow2x=5,36\)
\(\Leftrightarrow x=2,68\)
b) \(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2005.2010}\)
\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{2010}\right)\)
\(\Leftrightarrow N=\frac{1}{5}.\frac{2009}{2010}=\frac{2009}{10050}\)
Bài 1:
a)\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot42,9-11,7\cdot0,25+0,25\cdot0,8\)
\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot\left(42,9-11,7+0,8\right)\)
\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot32\)
\(\frac{2\cdot x-4,36}{0,125}=8\)
\(2\cdot x-4,36=8\cdot0,125\)
\(2\cdot x-4,36=1\)
\(2\cdot x=1+4,36\)
\(2\cdot x=5,36\)
\(x=\frac{5,36}{2}=2,68\)
b) \(N=\frac{1}{1\cdot5}+\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+\frac{1}{15\cdot20}+...+\frac{1}{2005\cdot2010}\)
\(4N=\frac{4}{1\cdot5}+\frac{4}{5\cdot10}+\frac{4}{10\cdot15}+\frac{4}{15\cdot20}+...+\frac{4}{2005\cdot2010}\)
\(4N=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\)
\(4N=1-\frac{1}{2010}=\frac{2009}{2010}\)
\(N=\frac{2009}{2010}\div4=\frac{2009}{8040}\)
Bài 2:
a) ( x + 5,2 ) : 3,2 = 4,7 ( dư 0,5 )
\(x+5,2=4,7\cdot3,2+0,5\)
\(x+5,2=15,54\)
\(x=15,54-5,2=10,34\)
b)\(A=\frac{4047991-2010\cdot2009}{4050000-2011\cdot2009}\)
\(A=\frac{4047991-2010\cdot2009}{4050000-2009-2010\cdot2009}\)
\(A=\frac{4047991-2010\cdot2009}{4047991-2010\cdot2009}=1\)
Bài 3:
a) \(104,5\cdot x-14,1\cdot x+9,6\cdot x=25\)
\(x\cdot\left(104,5-14,1+9,6\right)=25\)
\(x\cdot100=25\)
\(x=\frac{25}{100}=\frac{1}{4}=0,25\)
b) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)
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, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)
= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)
= \(\frac{1}{3}\)- \(\frac{1}{15}\)
= \(\frac{4}{15}\)