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Ta có:
\(\sqrt{2016}-\sqrt{2017}=\frac{\left(\sqrt{2016}-\sqrt{2017}\right)\left(\sqrt{2016}+\sqrt{2017}\right)}{\sqrt{2016}+\sqrt{2017}}\)
\(=\frac{2016-2017}{\sqrt{2016}+\sqrt{2017}}=-\frac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\sqrt{2017}-\sqrt{2018}=\frac{\left(\sqrt{2017}-\sqrt{2018}\right)\left(\sqrt{2017}+\sqrt{2018}\right)}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{2017-2018}{\sqrt{2017}+\sqrt{2018}}=-\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
Ta thấy rằng:
\(\sqrt{2018}>\sqrt{2016}\)
\(\Leftrightarrow\sqrt{2017}+\sqrt{2018}>\sqrt{2016}+\sqrt{2017}\)
\(\Leftrightarrow\frac{1}{\sqrt{2017}+\sqrt{2018}}< \frac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\Leftrightarrow-\frac{1}{\sqrt{2017}+\sqrt{2018}}>-\frac{1}{\sqrt{2016}+\sqrt{2017}}\)
Vậy \(\sqrt{2017}-\sqrt{2018}>\sqrt{2016}-\sqrt{2017}\)
Đặt \(A=\left(\sqrt{2018}+\sqrt{2020}\right)\)
\(\Rightarrow A^2=2018+2\sqrt{2018.2020}+2020=4038+\sqrt{4.2018.2020}=4038+\sqrt{4.\left(2019^2-1\right)}\)
Đặt \(B=2\sqrt{2019}=\sqrt{4.2019}\)
\(B^2=4.2019=2.2019+2.2019=4038+\sqrt{4.2019^2}\)
=> \(\sqrt{4.2019^2}>\sqrt{4.\left(2019^2-1\right)}\)
\(\Rightarrow A>B\Leftrightarrow\sqrt{2018}+\sqrt{2020}>2\sqrt{2019}\)
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
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 \(x>y\)
\(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2017}+\sqrt{2016}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
2017>2015
=>căn 2017>căn 2015
=>\(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}+\sqrt{2015}\)
=>\(\dfrac{1}{\sqrt{2017}+\sqrt{2016}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
=>\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
theo em là A=B
em mới học lớp 5 thôi chưa chắc đúng đâu
2017=2017
2018 hơn 2016 là 2 đơn vị
2017 lớn hơn 2016 là 1 đơn vị
2017 lớn hơn 2016 1 đơn vị
A hơn B số đăn vị là:
2-(1+1)=0
Nên A=B
thanks em nha anh sẽ xem lại
Ai có kết quả nữa thì giúp mình nha
Ta có: \(\left(\sqrt{2015}+\sqrt{2018}\right)^2=4033+2\sqrt{2015.2018}\)
\(\left(\sqrt{2016}+\sqrt{2017}\right)^2=4033+2\sqrt{2016.2017}\)
\(2015.2018=2015.2017+2015=2017\left(2015+1\right)-2017+2015=2017.2016-2\)\(\Rightarrow2015.2018< 2016.2017\)
\(\Rightarrow4033+2\sqrt{2015.2018}< 4033+2\sqrt{2016.2017}\)
\(\Rightarrow\sqrt{2015}+\sqrt{2018}< \sqrt{2016}+\sqrt{2017}\left(đpcm\right)\)
\(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2019+2\sqrt{2017.2019}\)
\(=2.2018+2\sqrt{2018^2-1}< 2.2018+2.2018=4.2018\)
Ta có: \(\left(\sqrt{2017}+\sqrt{2019}\right)^2< 4.2018\)
\(\Rightarrow\sqrt{2017}+\sqrt{2018}< 2.\sqrt{2018}\)
Tham khảo nhé~
Lời giải:
\(\sqrt{2019}-\sqrt{2018}=\frac{2019-2018}{\sqrt{2019}+\sqrt{2018}}=\frac{1}{\sqrt{2019}+\sqrt{2018}}\)
\(\sqrt{2018}-\sqrt{2017}=\frac{2018-2017}{\sqrt{2018}+\sqrt{2017}}=\frac{1}{\sqrt{2018}+\sqrt{2017}}\)
Dễ thấy \(\sqrt{2019}+\sqrt{2018}>\sqrt{2018}+\sqrt{2017}\Rightarrow \frac{1}{\sqrt{2019}+\sqrt{2018}}< \frac{1}{\sqrt{2018}+\sqrt{2017}}\)
\(\Rightarrow \sqrt{2019}-\sqrt{2018}< \sqrt{2018}-\sqrt{2017}\)