1) Chứng minh rằng: \(\frac{1}{2^2}+\frac{1}{3^3}\)\(+\frac{1}{4^4}\)\(+........+\frac{1}{1990^2}\)\(< \frac{3}{4}\)
2)Tính: \(\frac{1}{19}\)\(+\frac{9}{19.29}\)\(+\frac{9}{29.39}\)\(+........+\frac{9}{1999.2009}\)
K
Khách
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TH
1
9 tháng 9 2018
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)\)
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{2000}{18081}\right)\)
\(A=\frac{1}{19}+\frac{200}{2009}\)
\(A=\frac{5809}{38171}\)
MK ko chắc nhé =v ( mấy bước quy đồng lằng nhằng ko làm âu )
22 tháng 5 2017
\(A=\frac{50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}}{100-\frac{8}{13}+\frac{4}{15}-\frac{4}{17}}\)
\(=\frac{50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}}{2\left(50-\frac{4}{13}+\frac{2}{15}-\frac{2}{17}\right)}\)
\(=\frac{1}{2}\)
\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+......+\frac{9}{1999.2009}\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{19}-\frac{1}{39}+....+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\cdot\frac{1990}{38171}\)
\(=\frac{200}{2009}\)
27 tháng 12 2015
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)=\frac{1}{19}+\frac{9}{10}\cdot\frac{1990}{38171}=\frac{1}{19}+\frac{1791}{38171}=\frac{200}{2009}\)
G6
3
15 tháng 7 2018
\(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}.\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}.\left(\frac{1}{19}-\frac{1}{2009}\right)\)
b tự làm nốt nhé
15 tháng 7 2018
\(\frac{1}{9.19}+\frac{1}{19.29}+\frac{1}{29.39}+...+\frac{1}{1999.2009}\)
\(=\frac{1}{10}\times\left(\frac{10}{9.19}+\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(=\frac{1}{10}\times\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{10}\times\left(\frac{1}{9}-\frac{1}{2009}\right)\)
\(=\frac{1}{10}\times\frac{2000}{18081}\)
\(=\frac{200}{18081}\)
_Chúc bạn học tốt_
22 tháng 5 2017
Ta có:
\(A=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\) \(\dfrac{9}{1999.2009}\)
\(=\dfrac{1}{19}+\) \(\left(\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\right)\)
\(=\dfrac{1}{19}\) \(+\) \(\dfrac{9}{10}\left(\dfrac{10}{19.29}+\dfrac{10}{29.39}+...+\dfrac{10}{1999.2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{39}+...+\dfrac{1}{1999}-\dfrac{1}{2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)=\dfrac{200}{2009}\)
Vậy \(A=\dfrac{200}{2009}\)
1 tháng 3 2017
tr tốc độ kinh khủng!!!! ms có 4p mà p vừa đánh máy vừa suy nghĩ hả ? BÁI PHỤC !!!
23 tháng 2 2017
\(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
= \(\frac{1}{19}+\left(\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\right)\)
= \(\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
= \(\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
= \(\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
= \(\frac{1}{19}+\frac{9}{10}.\frac{1990}{38171}\)
= \(\frac{1}{19}+\frac{1791}{38171}\)
= \(\frac{200}{2009}\)
24 tháng 10 2016
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
DP
19 tháng 7 2017
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\text{(đpcm) }\)
1/
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{1990^2}< \frac{1}{1989.1990}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)
\(VP=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)
=> ĐPCM
2/
\(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29+39}+...+\frac{9}{1999.2009}\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}.\frac{1990}{38171}\)
\(=\frac{200}{2009}\)