\(2>\sqrt{5-3}\)
\(\sqrt{2}+\sqrt{3}>2\)
\(5-3\sqrt{2}>2\sqrt{3}-3\)
\(2-\sqrt{2}>\dfrac{1}{2}\)
a) Ta có:
\(\sqrt{5-3}=\sqrt{2}\)
\(2=\sqrt{4}\)
Vì 4 > 2 nên \(\sqrt{4}>\sqrt{2}\). \(\Rightarrow2>\sqrt{5-3}\)
b) Ta có: \(\sqrt{3}>\sqrt{2}>\sqrt{1}=1\Rightarrow\sqrt{3}+\sqrt{2}>1+1=2\)
Vậy, \(\sqrt{3}+\sqrt{2}>2\)
Ta có: \(\left\{{}\begin{matrix}3\sqrt{2}=\sqrt{9.2}=\sqrt{18}>\sqrt{16}=4\\2\sqrt{3}=\sqrt{4.3}=\sqrt{12}< \sqrt{16}=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5-3\sqrt{2}>5-4=1\\2\sqrt{3}-3< 4-3=1\end{matrix}\right.\)
\(\Rightarrow5-3\sqrt{2}>2\sqrt{3}-3\)
a) Vì 4 > 5 - 3
⇒\(\sqrt{4}\) > \(\sqrt{5-3}\)
⇒2 > \(\sqrt{5-3}\)
Vậy 2 > \(\sqrt{5-3}\)
b)Vì 2 > 1 và 3 > 1
⇒\(\sqrt{2}\) > 1 và \(\sqrt{3}\) > 1
⇒\(\sqrt{2}+\sqrt{3}>2\)
Vậy \(\sqrt{2}+\sqrt{3}>2\)