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29 tháng 3 2017

\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{17\cdot21}< 1\)

\(A=\dfrac{4}{4}\cdot\left(\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+...+\dfrac{1}{17\cdot21}\right)< 1\)

\(A=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{17}-\dfrac{1}{21}< 1\)

\(A=1-\dfrac{1}{21}< 1\) (đúng) (đpcm).

29 tháng 3 2017

Đề sai

21 tháng 4 2020

Chứng minh \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)

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

\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)

\(A=\frac{1}{1}-\frac{1}{21}\)

\(A=\frac{20}{21}\)

\(\frac{20}{21}< 1\)

=> \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)( đpcm ) 

* Mình sợ sai xD *

29 tháng 3 2017

Tìm x

\(\dfrac{x}{5}\)=\(\dfrac{x+6}{15}\)

\(\Rightarrow\)\(\dfrac{3x}{15}\)=\(\dfrac{x+6}{15}\)

\(\Rightarrow\)3x = x+6

\(\Rightarrow\)2x=6

\(\Rightarrow\)x=3

TÍNH TỔNG S

S=\(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{17.21}\)

S=\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{17}-\dfrac{1}{21}\)

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

S= \(\dfrac{20}{21}\)

29 tháng 3 2017

Tìm x:

\(\dfrac{x}{5}=\dfrac{x+6}{15}=>\dfrac{3x}{15}=\dfrac{x+6}{15}\)

=> 3x = 6 + x

=> 2x = 6

=> x = 3

Tính tổng S:

\(S=\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{17.21}\)

\(S=\dfrac{4}{1}-\dfrac{4}{5}+\dfrac{4}{5}-\dfrac{4}{9}+\dfrac{4}{9}-\dfrac{4}{13}+...+\dfrac{4}{17}-\dfrac{4}{21}\)

\(S=4-\dfrac{4}{21}\)

\(S=\dfrac{80}{21}\)

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 ....

23 tháng 7 2020

ủa, \(\frac{20}{\frac{21}{4}}\)t bấm máy tính đâu ra \(\frac{5}{21}\)đâu nhở?

24 tháng 7 2020

là 20/21 // 4 cơ

10 tháng 5 2017

\(A=8400\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(=\frac{8400}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\right)\)

\(=2100\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)

\(=2100\left(1-\frac{1}{25}\right)\)

\(=2100\cdot\frac{24}{25}\)

\(=2016\)

10 tháng 5 2017

\(A=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(A=8400.\left(\frac{1.4}{1.5.4}+\frac{1.4}{5.9.4}+\frac{1.4}{9.13.4}+\frac{1.4}{13.17.4}+\frac{1.4}{17.21.4}+\frac{1.4}{21.25.4}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{25}\right)\)

\(A=8400.\frac{1}{4}.\frac{24}{25}\)

\(A=2016\)