\(\dfrac{45.21.25}{7.15.36}=\dfrac{3.3.3.5.7.5.5}{7.5.3.3.2.3.2}=\dfrac{25}{4}\)

45\7 x 21\15 x 25/36

=45x21x25/7x15x36

=15x3x7x3x25/7x15x3x12

=3x25/12

=75/12

=25/4

7/15 nhé bạn!

đặt A=1/6+1/10+1/15+1/21+1/28+1/36+1/45

A*2=(1/6*+1/10+1/15+1/21+1/28+1/36+1/45)*2

A*2=1/12+1/20+1/30+1/42+1/56+1/72+1/90

A*2=1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9+1/9*10

A*2=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-/8+1/8-1/9+1/9-1/10

A*2=1/3-1/10

A*2=7/30

A=7/30 / 2

A=7/15

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

\(=2.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

\(=2.\left(\frac{1}{3}-\frac{1}{10}\right)\)

\(=2.\left(\frac{10}{30}-\frac{3}{30}\right)\)

\(=2.\frac{7}{30}\)

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

Chúc bạn học tốt !!! 

\(\frac{4}{5}\)

đặt A=1/6+1/10+1/15+1/21+1/28+1/36+1/45

6A=1+3/5+2/5+2/7+3/14+1/6+2/15

6A=1+1+7/14+1/6+2/15

6A=14/5

A=14/5:6=7/15

Đặt A = 1+1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45 
Nhân 2 vế với 1/2 để xuất hiện các mẫu là tích của 2 số tự nhiên liên tiếp sau đó áp dụng công thức 1/n.(n + 1) = 1/n - 1/(n + 1) ta có 
1/2.A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 
1/2.A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 
1/2.A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +.......+ 1/8 - 1/9 + 1/9 - 1/10 
1/2.A = 1- 1/10 
1/2.A = 9/10 
=> A = 9/5 

Đặt A = 1+1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45

Nhân 2 vế với 1/2 để xuất hiện các mẫu là tích của 2 số tự nhiên liên tiếp sau đó áp dụng công thức 1/n.(n + 1) = 1/n - 1/(n + 1) ta có :

1/2.A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90

1/2.A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10

1/2.A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +.......+ 1/8 - 1/9 + 1/9 - 1/10

1/2.A = 1- 1/10

1/2.A = 9/10

=> A = 9/5 

Đặt A = 1+1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45

Nhân 2 vế với 1/2 để xuất hiện các mẫu là tích của 2 số tự nhiên liên tiếp sau đó áp dụng công thức 1/n.(n + 1) = 1/n - 1/(n + 1) ta có :

 1/2.A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 

1/2.A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10

1/2.A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +.......+ 1/8 - 1/9 + 1/9 - 1/10

1/2.A = 1- 1/10

1/2.A=9/10

=> A = 9/5 

ta có

(1/3+1/6+1/36) +(1/10+1/15+1/45)+(1/21+1/28)

=(\(\frac{12+6+1}{36}\)+\(\frac{9+6+2}{90}\)+\(\frac{4+3}{84}\)

19/36+17/90+1/12

=(19/36+1/12)+17/90

=7/12+17/90

=105/180+34/180

=139/180

1/3 +1/6+1/10+1/15+1/21+1/28+1/36+1/45

=1/1x3+1/3x2+1/2x5+1/3x5+1/3x7+1/7x4+1/4x9+1/9x5

=1/1-1/3+1/3-1/2....+1/9-1/5

=1/1

bạn ơ sao mình tính trong máy tính đc 4/5 mà

\(P=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\)

\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=2.\dfrac{2}{5}=\dfrac{4}{5}\)

bn cộng hết lại vs nhau là ra kq!

1 / 6 + 1 / 10  + 1 / 15 + 1 / 21 +  1 / 28 + 1 / 36 + 1 / 45

= 2 / 12 + 2 / 10 + 2 / 30 + 2 / 42  + 2 / 56 + 2 / 72 + 2 / 90

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

= 2 .( 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)

=2 ( 1 / 3 - 1 / 10 )

= 2. 7 / 30

= 7 / 15

Bài làm:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ta có:

\(A=\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{210}\)

=> \(\dfrac{1}{2}A=\dfrac{1}{2}\left(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{210}\right)\text{​}\)

\(=\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{420}\)

\(=\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{20.21}\)

\(=\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{20}-\dfrac{1}{21}\)

\(=\dfrac{1}{6}-\dfrac{1}{21}\)

\(=\dfrac{5}{42}\)

Vậy \(A=\dfrac{5}{42}\)