Question

So sánh

\(\dfrac{2019}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2019}}\) và \(\sqrt{2018}+\sqrt{2019}\)

Answer 1

Ta có:

\(\dfrac{2019}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2019}}=\dfrac{2018}{\sqrt{2018}}+\dfrac{1}{\sqrt{2018}}+\dfrac{2019}{\sqrt{2019}}-\dfrac{1}{\sqrt{2019}}=\sqrt{2018}+\sqrt{2019}+\left(\dfrac{1}{\sqrt{2018}}-\dfrac{1}{\sqrt{2019}}\right)\)

Do \(\dfrac{1}{\sqrt{2018}}>\dfrac{1}{\sqrt{2019}}\) nên \(\dfrac{1}{\sqrt{2018}}-\dfrac{1}{\sqrt{2019}}\) dương \(\Rightarrow\dfrac{2019}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2019}}>\sqrt{2018}+\sqrt{2019}\)

Answer 2

20192018−−−−√+20182019−−−−√=20182018−−−−√+12018−−−−√+20192019−−−−√−12019−−−−√=2018−−−−√+2019−−−−√+(12018−−−−√−12019−−−−√)20192018+20182019=20182018+12018+20192019−12019=2018+2019+(12018−12019)

Do 12018−−−−√>12019−−−−√12018>12019 nên 12018−−−−√−12019−−−−√12018−12019 dương ⇒20192018−−−−√+20182019−−−−√>2018−−−−√+2019−−−−√

Answer 3

Ta có : 2019>2018

<=>\(\sqrt{2019}\) > \(\sqrt{2019}\)

<=> \(\dfrac{1}{\sqrt{2019}}\) < \(\dfrac{1}{\sqrt{2018}}\)

<=> \(\dfrac{2019-2018}{\sqrt{2019}}\) < \(\dfrac{2019-2018}{\sqrt{2018}}\)

<=> \(\sqrt{2019}-\dfrac{2018}{\sqrt{2019}}\) <\(\dfrac{2019}{\sqrt{2018}}-\sqrt{2018}\)

<=> \(\sqrt{2018}+\sqrt{2019}\) <\(\dfrac{2019}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2019}}\)