sao sánh hộ: A=√2019 +√2017 và B=2√2018
Cách khác :vv
$A-B=(\sqrt{2019}+\sqrt{2017})-(2\sqrt{2018})=(\sqrt{2019}-\sqrt{2018})-(\sqrt{2018}-\sqrt{2017})=\frac{1}{\sqrt{2019}+\sqrt{2018}}-\frac{1}{\sqrt{2018}+\sqrt{2017}}$
Có $\sqrt{2019}+\sqrt{2018}>\sqrt{2018}+\sqrt{2017}$
$\Rightarrow \dfrac{1}{\sqrt{2019}+\sqrt{2018}}<\dfrac{1}{\sqrt{2018}+\sqrt{2017}}$
$\Rightarrow A-B<0$ hay $A<B$