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tran thu thuy
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sakura
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DORAEMON
24 tháng 4 2016 lúc 16:44

P=2(1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)

P=2((1/2*3)+(1/3*4)+(1/4*5)+(1/5*6)+(1/6*7)+(1/7*8)+(1/8*9)+(1/9*10)

P=2(1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)

P=2(1/2-1/10)

P=2*2/5

P=4/5

H NHA.

chu cẩm linh
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Nguyễn Tuấn Dũng
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nguyen min0h hoang
11 tháng 2 2017 lúc 22:42

ta có 
1/2 P=1/2(1-1/10-1/15-1/3-1/28-1/6-1/21)
=1/2-(1/6+1/12+1/20+1/30+1/42+1/56)
=1/2-(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8)
=1/2-(1/2-1/8)
=1/8
suy ra P=1/4

nguyen min0h hoang
11 tháng 2 2017 lúc 22:43

ta có 
1/2 P=1/2(1-1/10-1/15-1/3-1/28-1/6-1/21)
=1/2-(1/6+1/12+1/20+1/30+1/42+1/56)
=1/2-(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8)
=1/2-(1/2-1/8)
=1/8
suy ra P=1/4

trườngkute
30 tháng 10 2017 lúc 4:31

A=/x-2/+3x+2

Hồ Thanh Vân
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Vũ Thị Minh Nguyệt
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Huỳnh Huyền Linh
2 tháng 4 2017 lúc 14:51

21)

\(\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{8}\right).\left(1+\dfrac{1}{15}\right).....\left(1+\dfrac{1}{9999}\right)\\ =\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{10000}{9999}\\ =\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{100.100}{99.101}\\ =\dfrac{2.3.4.....100}{1.2.3.....99}.\dfrac{2.3.4.....100}{3.4.5.....101}\\ =100.\dfrac{2}{101}\\ =\dfrac{200}{101}\)

Nguyễn Văn Vi Duy Hưng
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a: \(A=\left(\frac{136}{15}-\frac{28}{5}+\frac{62}{10}\right)\cdot\frac{21}{24}\)

\(=\left(\frac{272}{30}-\frac{168}{30}+\frac{186}{30}\right)\cdot\frac78\)

\(=\frac{290}{30}\cdot\frac78=\frac{29}{3}\cdot\frac78=\frac{203}{24}\)

b: \(B=\frac56+6\frac56\left(11\frac{5}{20}-9\frac14\right):8\frac13\)

\(=\frac56+\frac{41}{6}\cdot\left(11+\frac14-9-\frac14\right):\frac{25}{3}=\frac56+\frac{41}{6}\cdot2\cdot\frac{3}{25}\)

\(=\frac56+\frac{41}{25}=\frac{125}{150}+\frac{246}{150}=\frac{371}{150}\)

c: \(C=1+3+6+\cdots+1225\)

\(=\frac12\left(2+6+12+\cdots+2450\right)=\frac12\cdot\left(1\cdot2+2\cdot3+\cdots+49\cdot50\right)\)

\(=\frac12\cdot\left\lbrack1\left(1+1\right)+2\left(2+1\right)+\cdots+49\left(49+1\right)\right\rbrack=\frac12\cdot\left\lbrack\left(1+2+\cdots+49\right)+\left(1^2+2^2+\cdots+49^2\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack49\cdot\frac{50}{2}+\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack\)

\(=\frac12\cdot\left\lbrack49\cdot25+49\cdot50\cdot\frac{99}{6}\right\rbrack=\frac12\cdot\left\lbrack49\cdot25+49\cdot25\cdot33\right\rbrack=\frac12\cdot49\cdot25\cdot\left(33+1\right)\)

\(=49\cdot25\cdot\frac{34}{2}=49\cdot25\cdot17=20825\)

Đỗ Hoàng Minh
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A =          1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) +  \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\)

A = 2\(\times\) ( \(\dfrac{1}{2}\)  +  \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) +  \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\)\(\dfrac{1}{72}\))

A =2\(\times\)\(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\))

A = 2 \(\times\) ( \(\dfrac{1}{1}\)\(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\))

 A = 2\(\times\)( 1 - \(\dfrac{1}{9}\))

A = 2 \(\times\) \(\dfrac{8}{9}\)

A = \(\dfrac{16}{9}\)

Linh Chi
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Nguyễn Minh Đăng
8 tháng 8 2020 lúc 20:19

Bài làm:

Ta có: \(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{66}\)

\(=\frac{1}{1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{11.6}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.1.3}+\frac{1}{2.3.2}+...+\frac{1}{2.11.6}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{12}\right)\)

\(=\frac{1}{2}.\frac{11}{12}\)

\(=\frac{11}{24}\)

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Ngoc Han ♪
8 tháng 8 2020 lúc 20:21

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)

\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)

\(=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{10\times11}+\frac{1}{11\times12}\right)\)

\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)

\(=2\times\left(1-\frac{1}{12}\right)\)

\(=2\times\frac{11}{12}\)

\(=\frac{11}{6}\)

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ミ★Ƙαї★彡
8 tháng 8 2020 lúc 20:38

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)

\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)

\(=2\left(1-\frac{1}{12}\right)=2.\frac{11}{12}=\frac{22}{12}=\frac{11}{6}\)

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