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\(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}\)
\(\Rightarrow A< \frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{2018}=\frac{1009}{2018}-\frac{1}{2018}\)
\(\Rightarrow A< \frac{1008}{2018}=\frac{504}{1009}\)
\(\Rightarrow\) \(A< \frac{504}{1009}\left(đpcm\right)\)
Dòng thứ 2 sao lại : \(A< \frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)vậy bạn
2.4 ở đâu
Làm bài 1 thui nhé, mấy bài kia dễ tự làm -,-
\(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}\)
\(A< \frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{1}{2}\left(\frac{1}{2}-\frac{2}{2017}\right)< \frac{1}{2}\left(\frac{1}{2}-\frac{2}{2018}\right)=\frac{1}{2}.\frac{1007}{2018}\)
\(\Rightarrow\)\(2A< \frac{1007}{2018}< \frac{1008}{2018}=\frac{504}{1009}\)\(\Rightarrow\)\(A< \frac{504}{1009}\)
Vậy \(A< \frac{504}{1009}\)
Chúc bạn học tốt ~
Ta có : M = \(\frac{x+y}{z}+\frac{x+z}{y}=\frac{y+z}{x}\)
\(\Rightarrow M+3=\left(\frac{x+y}{z}+1\right)+\left(\frac{x+z}{y}+1\right)+\left(\frac{y+z}{x}+1\right)\)
\(\Rightarrow M+3=\frac{x+y+z}{z}+\frac{x+y+z}{y}+\frac{x+y+z}{x}\)
\(\Rightarrow M+3=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\Rightarrow M+3=2020.\frac{1}{202}\)
=> M + 3 = 10
=> M = 7
Vậy M = 7
b) Ta có : \(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}\)
\(=\frac{2}{3.3}+\frac{2}{5.5}+\frac{2}{7.7}+...+\frac{2}{2017.2017}\)
\(< \frac{2}{\left(3+1\right)\left(3-1\right)}+\frac{2}{\left(5-1\right)\left(5+1\right)}+\frac{2}{\left(7-1\right)\left(7+1\right)}+...+\frac{2}{\left(2017-1\right)\left(2016-1\right)}\)
\(=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}\)
\(=\frac{1008}{2018}=\frac{504}{1009}\)
=> \(A< \frac{504}{1009}\left(\text{ĐPCM}\right)\)
Lời giải:
Xét số hạng tổng quát: \(\frac{2}{(2n+1)^2}\)
Thấy rằng $(2n+1)^2=4n^2+4n+1>4n^2+4n=2n(2n+2)$
$\Rightarrow \frac{2}{(2n+1)^2}< \frac{2}{2n(2n+2)}$
Cho $n=1,2,3...$ ta có:
$\frac{2}{3^2}< \frac{2}{2.4}$
$\frac{2}{5^2}< \frac{2}{4.6}$
....
$\frac{2}{2017^2}< \frac{2}{2016.2018}$
Cộng theo vế:
$\Rightarrow A< \frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2016.2018}$
$\Leftrightarrow A< \frac{4-2}{2.4}+\frac{6-4}{4.6}+....+\frac{2018-2016}{2016.2018}$
$\Leftrightarrow A< \frac{1}{2}-\frac{1}{2018}$
$\Leftrightarrow A< \frac{504}{1009}$
Ta có đpcm.
\(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}< \frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2015\cdot2017}\\ =1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\\ =1-\frac{1}{2017}=\frac{2016}{2017}>\frac{504}{1009}\)
Đề vô lí quá bạn ạ! Bạn xem lại đề giúp mình , có thể mình làm sai!
<3 <3 <3
\(A=2\cdot\left(\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2017^2}\right)< 2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}\right)\)
Đặt \(M=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}=\left(1+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(\Rightarrow M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(\Rightarrow M=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}< \frac{1}{1009}+\frac{1}{1009}+...+\frac{1}{1009}\)(1008 số hạng )
hay\(M< \frac{1008}{1009}\Rightarrow A< 2\cdot\frac{1008}{1009}=\frac{504}{1009}\left(ĐPCM\right)\)