Chứng minh:
\(\sqrt{2017}+\sqrt{2018}< \dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}\)
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Những câu hỏi liên quan
8 tháng 10 2021
Áp dụng BĐT Cauchy–Schwarz ta được:
\(x=\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}\ge\dfrac{\left(\sqrt{2018}+\sqrt{2017}\right)^2}{\sqrt{2018}+\sqrt{2017}}=\sqrt{2018}+\sqrt{2017}=y\)
Dấu \("="\Leftrightarrow\dfrac{2017}{\sqrt{2018}}=\dfrac{2018}{\sqrt{2017}}\Leftrightarrow2017=2018\left(vô.lí\right)\)
Vậy đẳng thức ko xảy ra hay \(x>y\)
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 \(\frac{\sqrt{2017}}{\sqrt{2018}}+\frac{\sqrt{2018}}{\sqrt{2017}}\ge2\Rightarrow\frac{\sqrt{2017}}{\sqrt{2018}}+\frac{\sqrt{2018}}{\sqrt{2017}}-1\ge1\)
\(\Rightarrow\sqrt{2017}+\sqrt{2018} < \frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}\)
AS
0
DN
2
TD
16 tháng 6 2018
Đat 2017,5=t Ta có
\(\sqrt{\dfrac{\left(t+0,5\right)^2+\left(t-0,5\right)^2\cdot\left(t+0,5\right)^2+\left(t-0,5\right)^2}{\left(t+0,5\right)^2}}+\dfrac{t-0,5}{t+0,5}\\ =\sqrt{\dfrac{t^2+t+0,25+t^4-0,5t^2+0,0625+t^2-t+0,25}{\left(t+0,5\right)^2}}+\dfrac{t-0,5}{t+0,5}\\ =\dfrac{\sqrt{t^4+1,5t^2+0,5625}}{t+0,5}+\dfrac{t-0,5}{t+0,5}\\ =\dfrac{t^2+0,75+t-0,5}{t+0,5}\\ =\dfrac{\left(t+0,5\right)^2}{t+0,5}\\ =t+0,5\)thay t=2017,5 vào suy ra A=2017,5+0,5=2018
HH
16 tháng 6 2018
Giải:
\(\sqrt{1+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
\(=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{\left(\dfrac{1}{2017}\right)^2}+\dfrac{1}{\left(-\dfrac{2018}{2017}\right)^2}}+\dfrac{2017}{2018}\)
\(=\sqrt{\left(\dfrac{1}{1}+\dfrac{1}{\dfrac{1}{2017}}+\dfrac{1}{-\dfrac{2018}{2017}}\right)^2}+\dfrac{2017}{2018}\) (\(\left\{{}\begin{matrix}1>0\\2017^2>0\\\dfrac{2017^2}{2018^2}>0\end{matrix}\right.\Leftrightarrow1+2017^2+\dfrac{2017^2}{2018^2}>0\ne0\))
\(=1+2017+-\dfrac{2017}{2018}+\dfrac{2017}{2018}\)
\(=2018\)
Vậy ...
14 tháng 10 2021
Đặt \(2017=a\)
\(A=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2a+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2\left(a+1\right)\cdot\dfrac{a}{a+1}+\left(\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1-\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\left|a+1-\dfrac{a}{a+1}\right|+\dfrac{a}{a+1}\)
Ta có \(\dfrac{a}{a+1}< 1\Leftrightarrow a+1-\dfrac{a}{a+1}>0\)
\(\Leftrightarrow A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2018\)
TT
1
31 tháng 5 2022
Sửa đề: \(M=\sqrt{1^2+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
\(=\sqrt{1^2+\dfrac{1}{\left(\dfrac{1}{2017}\right)^2}+\dfrac{1}{\left(-\dfrac{2018}{2017}\right)^2}}+\dfrac{2017}{2018}\)
\(=\sqrt{\left(\dfrac{1}{1}+\dfrac{1}{\dfrac{1}{2017}}+\dfrac{1}{-\dfrac{2018}{2017}}\right)^2}+\dfrac{2017}{2018}\)
\(=1+2017-\dfrac{2017}{2018}+\dfrac{2017}{2018}\)
=2018
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 \(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}>\sqrt{2017}+\sqrt{2018}\)
không thể cm