![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
2016.x2017 + 2018 = 4034
2016.x2017 = 4034-2018
2016.x2017 = 2016
x2017 = 1
=>x =1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{2015}{2016}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{2016}-\frac{1}{2017}\)
\(=\frac{1}{2}-\frac{1}{2017}=\frac{2015}{4024}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}>\frac{2015}{4034}\)
vậy ta có điều cần chứng minh
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\) ta có :
\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A>\frac{1}{2}-\frac{1}{2017}\)
\(A>\frac{2015}{4034}\) \(\left(1\right)\)
Lại có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A< 1-\frac{1}{2016}\)
\(A< \frac{2015}{2016}\) \(\left(2\right)\)
Từ (1) và (2) suy ra : \(\frac{2015}{4034}< A< \frac{2015}{2016}\) ( đpcm )
Vậy \(\frac{2015}{4034}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< \frac{2015}{2016}\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2.3 +1/3.4+...+1/2016.2017 < 1/2^2+1/3^2+...+1/2016^2
1/2 -1/3+1/3 -1/4+...+1/2016-1/2017 < 1/2^2+1/3^2+...+1/2016^2
1/2-1/2017 < 1/2^2+1/3^2+...+1/2016^2
=> 2015/4034 < 1/2^2+1/3^2+...+1/2016^2
tương tự
1/2^2+1/3^2+...+1/2016^2 < 1/1.2 +1/2.3+...+1/2015.2016
1/2^2+1/3^2+...+1/2016^2 < 1- 1/2+1/2 -1/3+...+1/2015- 1/2016
1/2^2+1/3^2+...+1/2016^2 < 1-1/2016
1/2^2+1/3^2+...+1/2016^2 < 2015/2016
tích nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)=\frac{4034}{5}\)
\(x.\frac{2}{5}=\frac{4034}{5}\)
\(x=\frac{4034}{5}:\frac{2}{5}\)
\(x=\frac{4034}{5}.\frac{5}{2}\)
x = 2017
![](https://rs.olm.vn/images/avt/0.png?1311)
Bấm vào đúng 0 sẽ hiện ra kết quả đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2019.2020-4040}{2017.2018+4034}\)=\(\frac{\left(2017+2\right).2020-4040}{2017.2018+2017.2}\)
=\(\frac{2017.2020+2.2020-4040}{2017.\left(2018+2\right)}\)
=\(\frac{2017.2020+4040-4040}{2017.2020}\)
=\(\frac{2017.2020+0}{2017.2020}\)
=\(\frac{1}{1}\)=1
\(\frac{2016\times2017+4034}{2018\times2019-4034}=\frac{2016\times2017+2\times2017}{2018\times2019-2\times2017}\)
\(=\frac{\left(2016+2\right)\times2017}{2018\times2017+2\times2017-2\times2017}=\frac{\left(2016+2\right)\times2017}{2018\times2017+0}\)
\(=\frac{2018\times2017}{2018\times2017}=1\)