So sánh x và y trong các TH sau: \(x=\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}};y=\sqrt{2017}+\sqrt{2018}\)
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
1 tháng 8 2018
a) Ta có: \(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2019+2\sqrt{2017.2019}\)
\(=4036+2\sqrt{\left(2018-1\right).\left(2018+1\right)}\)
\(=4036+2\sqrt{2018^2-1}< 4036+2\sqrt{2018^2}=2018.4=\left(2\sqrt{2018}\right)^2\)
Vậy x < y
MS
21 tháng 7 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 \(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}>\sqrt{2017}+\sqrt{2018}\)
KH
1
HP
19 tháng 1 2021
Áp dụng BĐT Cosi cho 2018 số:
\(2017.6^{2018}.\sqrt[2017]{m}+\dfrac{\left(2a\right)^{2018}}{m}\ge2018\sqrt[2018]{\left(6^{2018}.\sqrt[2017]{m}\right)^{2017}\dfrac{\left(2a\right)^{2018}}{m}}=2018.2.6^{2017}.a\)
\(\Leftrightarrow\dfrac{\left(2a\right)^{2018}}{m}\ge2018.2.6^{2017}.a-2017.6^{2018}.\sqrt[2017]{m}\)
\(\Leftrightarrow\dfrac{2\left(2a\right)^{2018}}{m}\ge2018.4.6^{2017}.a-2017.2.6^{2018}.\sqrt[2017]{m}\)
Tương tự: \(\dfrac{2\left(2b\right)^{2018}}{n}\ge2018.4.6^{2017}.b-2017.2.6^{2018}.\sqrt[2017]{n}\)
\(\dfrac{3.c^{2018}}{p}\ge2018.3.6^{2017}.c-2017.6^{2018}.3.\sqrt[2017]{p}\)
\(\Rightarrow S\ge2018.6^{2017}\left(4a+4b+3c\right)-2017.6^{2018}\left(2\sqrt[2017]{m}+2\sqrt[2017]{n}+3\sqrt[2017]{p}\right)\)
\(\ge2018.6^{2017}.42-2017.6^{2018}.7=7.6^{2018}>6^{2018}\)
Vậy \(S>6^{2018}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1 2021
a. ĐKXĐ: \(x\ge-1\)
\(y=\sqrt{x^3+1+2\sqrt{x^3+1}+1}+\sqrt{x^3+1-2\sqrt{x^3+1}+1}\)
\(=\sqrt{\left(\sqrt{x^3+1}+1\right)^2}+\sqrt{\left(\sqrt{x^3+1}-1\right)^2}\)
\(=\left|\sqrt{x^3+1}+1\right|+\left|1-\sqrt{x^3+1}\right|\ge\left|\sqrt{x^3+1}+1+1-\sqrt{x^3+1}\right|=2\)
b.
\(f\left(x\right)=\dfrac{x-1}{2}+\dfrac{2}{x-1}+\dfrac{1}{2}\ge2\sqrt{\dfrac{2\left(x-1\right)}{2\left(x-1\right)}}+\dfrac{1}{2}=\dfrac{5}{2}\)
c.
\(y=\dfrac{x-2018+1}{\sqrt{x-2018}}=\sqrt{x-2018}+\dfrac{1}{\sqrt{x-2018}}\ge2\sqrt{\dfrac{\sqrt{x-2018}}{\sqrt{x-2018}}}=2\)
HN
31 tháng 7 2018
a/ Ta có:
\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
31 tháng 7 2018
a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
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 ...
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 \(x>y\)