\(\text{Chứng minh rằng:}2017< \sqrt{\frac{2}{1}}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+...+\sqrt[2018]{\frac{2018}{2017}}< 2018\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
LH
22 tháng 5 2020
Câu b đề sai nha, bây giờ đặt \(a=\sqrt{2017},b=\sqrt{2018}\)
Ta có \(\frac{a^2}{b}+\frac{b^2}{a}< a+b\Leftrightarrow ab\left(\frac{a^2}{b}+\frac{b^2}{a}\right)< ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3< ab\left(a+b\right)\)(1)
Mà \(ab\left(a+b\right)\le\left(a^2-ab+b^2\right)\left(a+b\right)=a^3+b^3\)(2)
Từ (1), (2) => Sai
22 tháng 5 2020
a) Ta có:
\(\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{k+1-k}{\left(k+1\right)\sqrt{k}}=\frac{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}\)\(< \frac{2\sqrt{k+1}\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k+1}\sqrt{k}}=\frac{2}{\sqrt{k}}-\frac{2}{\sqrt{k+1}}\)
Cho k=1,2,....,n rồi cộng từng vế ta có:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{\left(n+1\right)\sqrt{n}}< \left(\frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}\right)+\left(\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}\right)\)\(+\left(\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}\right)+....+\left(\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n-1}}< 2\)
7 tháng 3 2018
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}\)
\(=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Từ đây ta có
\(VT< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2017}}-\frac{1}{\sqrt{2018}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2018}}\right)< 2\)
T
8 tháng 3 2018
Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}\)
\(\Leftrightarrow\sqrt{n}\left(\frac{1}{n}-\frac{1}{n1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\). Mà:
\(\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Từ đó, ta có:
\(VT< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2017}}-\frac{1}{\sqrt{2018}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2018}}\right)< 2\) (ĐPCM)
CY
1
21 tháng 1 2020
Áp dụng bđt Svacxo ta có :
\(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}\ge\dfrac{\left(\sqrt{2017}+\sqrt{2018}\right)^2}{\sqrt{2018}+\sqrt{2017}}=\sqrt{2017}+\sqrt{2018}\)
Dấu bằng xảy ra khi:
\(\dfrac{2017}{\sqrt{2018}}=\dfrac{2018}{\sqrt{2017}}\Leftrightarrow2017=2018\left(vl\right)\)
Suy ra không xảy ra dấu bằng
Vậy \(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}>\sqrt{2017}+\sqrt{2018}\)
CC
1
BH
27 tháng 7 2019
\(\frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}=\frac{2017\sqrt{2017}+2018\sqrt{2018}}{\sqrt{2017}\cdot\sqrt{2018}}\)
\(=\left(\sqrt{2017}+\sqrt{2018}\right)\cdot\frac{2017+2018-\sqrt{2018\cdot2017}}{\sqrt{2017\cdot2018}}\)
Ta thấy \(\frac{2017+2018-\sqrt{2018\cdot2017}}{\sqrt{2018\cdot2017}}=\frac{\sqrt{2017}}{\sqrt{2018}}+\frac{\sqrt{2018}}{\sqrt{2017}}-1\)
Áp dụng ĐBT Cô si thì \(\frac{\sqrt{2017}}{\sqrt{2018}}+\frac{\sqrt{2018}}{\sqrt{2017}}\ge2\Rightarrow\frac{\sqrt{2017}}{\sqrt{2018}}+\frac{\sqrt{2018}}{\sqrt{2017}}-1\ge1\)
\(\Rightarrow\sqrt{2017}+\sqrt{2018} < \frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}\)
30 tháng 7 2019
Gọi vế trái BPT là A.
Xét biểu thức tổng quát:
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{\left[n\left(n+1\right)\right]^2}}\\ =\frac{\sqrt{n^2\left(n^2+2n+1\right)+n^2+2n+1+n^2}}{n\left(n+1\right)}\\ =\frac{\sqrt{n^4+2n^3+3n^2+2n+1}}{n\left(n+1\right)}\\ =\frac{\sqrt{\left(n^2+n+1\right)^2}}{n\left(n+1\right)}\\ =\frac{n^2+n+1}{n\left(n+1\right)}\\ =\frac{n\left(n+1\right)+n+1-n}{n\left(n+1\right)}\\ =1+\frac{1}{n}-\frac{1}{n+1}\)
Suy ra:
\(A=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+1+...+1\right)+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\right)\) (2018 số hạng 1)
\(=2018+\frac{1}{2}-\frac{1}{2018}< 2018\)
Vậy \(A< 2018\left(đpcm\right)\).
Chúc bạn học tốt nha
.
30 tháng 7 2019
cảm ơn bạn nhé, mình đag ko bt cách chứng minh biểu thức tổng quát ;)
27 tháng 6 2018
\(=\frac{\sqrt{2}-1}{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{4}-\sqrt{3}\right)}+...+\frac{\sqrt{2018}-\sqrt{2017}}{\left(\sqrt{2017}+\sqrt{2018}\right)\left(\sqrt{2018}-\sqrt{2017}\right)}\)
\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{2018}-\sqrt{2017}}{2018-2017}\)
\(=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{3}}{1}+...+\frac{\sqrt{2018}-\sqrt{2017}}{1}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2018}-\sqrt{2017}=\sqrt{2018}-1\)
27 tháng 6 2018
\(=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=-\sqrt{1}+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2018}\)
\(=-\left(\sqrt{1}+\sqrt{2018}\right)\)