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17 tháng 7 2015

A= 1/5.9+1/9.13+1/13.17+1/17.21+1/21.25

4A= 4/5.9+4/9.13+4/13.17+4/17.21+4/21.25

4A= (1/5-1/9)+(1/9-1/13)+(1/13-1/17)+(1/17-1/21)+(1/21-1/25)

4A= 1/5- 1/25

4A= 4/25

A= 4/25 :4 

A= 1/25

13 tháng 7 2018

bài này nâng cao lớp 6 mk giải rồi bạn nhờ ai giảng hộ nha nếu bn 5 lên 6

B=1/4.(4/1.5+4/5.9+......+4/25.29)

B=1/4.(1-1/5+1/5-1/9+.....+1/25-1/29)

B=1/4.(1-1/29)

B=1/4.28/29

B=7/29

13 tháng 7 2018

\(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}+\frac{1}{25.29}\)

\(\Rightarrow4B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}+\frac{4}{25.29}\)

\(\Rightarrow4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}+\frac{1}{25}-\frac{1}{29}\)

\(\Rightarrow4B=1-\frac{1}{29}\)

\(\Rightarrow4B=\frac{29}{29}-\frac{1}{29}=\frac{28}{29}\)

\(\Rightarrow B=\frac{28}{29}:4=\frac{28}{29}.\frac{1}{4}=\frac{7}{29}\)

Vậy ....

15 tháng 8 2018

\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)

\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\frac{44}{45}\)

\(=\frac{11}{45}\)

15 tháng 8 2018

Đặt \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)  là A.

Ta có:

\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)

\(4A=4\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\right)\)

\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\)

\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\)

\(4A=1-\frac{1}{45}\)

\(4A=\frac{44}{45}\)

\(A=\frac{44}{45}:4\)

\(A=\frac{11}{45}\)

Vậy \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}=\frac{11}{45}\)

9 tháng 8 2015

a) A = 4/5.9 + 4/9.13 + 4/13.17 + ... + 4/41/45

A = 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + ... + 1/41 - 1/45

A = 1/5 - 1/45

A = 8/45

b) B = ( 1 - 1/2 ) . ( 1 - 1/3 ) . ( 1 - 1/4 ) . ..... . ( 1 - 1/100 )

B = 1/2 . 2/3 . 3/4 . .... . 99/100

B = \(\frac{1.2.3.......99}{2.3.4......100}\)

B = 1/100

9 tháng 8 2015

B = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)

B = \(\frac{1}{2}.\frac{2}{3}.....\frac{99}{100}\)

B = \(\frac{1}{100}\)

2 tháng 8 2019

\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+\frac{7}{13.17}+\frac{7}{17.21}\)

\(=\frac{7}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}\right)\)

\(=\frac{7}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)\)

\(=\frac{7}{4}.\left(1-\frac{1}{21}\right)\)

\(=\frac{7}{4}.\frac{20}{21}=\frac{7.4.5}{4.7.3}\)

\(=\frac{5}{3}\)

~ Rất vui vì giúp đc bn ~

2 tháng 8 2019

Bài giải

\(\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\)

\(=\frac{7}{4}\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\right)\)

\(=\frac{7}{4}\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)\)

\(=\frac{7}{4}\left(1-\frac{1}{21}\right)\)

\(=\frac{7}{4}\cdot\frac{20}{21}\)

\(=\frac{35}{21}\)

15 tháng 7 2018

Đề?

\(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)

\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\frac{8}{45}=\frac{2}{45}\)

15 tháng 7 2018

sai đề

1/5.9+1/9.13+1/13.17+ ...+1/41.45

dễ cái này em nhân 4 lên rồi tính như bình thường là đcj

11 tháng 7 2016

\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{97.101}\)

\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{4}.\frac{100}{101}\)

\(=\frac{25}{101}\)

7 tháng 8 2016

1/1.5 + 1/5.9 + 1/9.13 + ... + 1/97.101

= 1/4.(4/1.5 + 4/5.9 + 4/9.13 + ... + 4/97.101)

= 1/4.(1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/97 - 1/101)

= 1/4.(1 - 1/101)

= 1/4.100/101

= 25/101

7 tháng 8 2016

\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+........+\frac{1}{97.101}\)

\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+........+\frac{1}{97}-\frac{1}{101}\right)\)

\(=\frac{1}{4}\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{4}.\frac{100}{101}=\frac{25}{101}\)

2 tháng 7 2018

\(A=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+........+\frac{1}{100.104}\)

\(=\frac{1}{4}.\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+.......+\frac{4}{100.104}\right)\)

\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+.......+\frac{1}{100}-\frac{1}{104}\right)\)

\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{104}\right)\)

\(=\frac{1}{4}.\frac{99}{520}=\frac{99}{2080}\)

11 tháng 11 2018

\(H=\frac{2\cdot2}{1\cdot5}+\frac{2\cdot2}{5\cdot9}+...+\frac{2\cdot2}{45.49}\)

\(H=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{45\cdot49}\)

\(H=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{45}-\frac{1}{49}\)

\(H=1-\frac{1}{49}\)

\(H=\frac{48}{49}\)

\(H=\frac{2.2}{1.5}+\frac{2.2}{5.9}+\frac{2.2}{9.13}+...+\frac{2.2}{45.49}\)

\(\Rightarrow H=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{45.49}\)

\(\Rightarrow H=\frac{5-1}{1.5}+\frac{9-5}{5.9}+...+\frac{49-45}{45.49}\)

\(\Rightarrow H=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{45}-\frac{1}{49}\)

\(\Rightarrow H=1-\frac{1}{49}=\frac{48}{49}\)