c.
(\sqrt{5}-\sqrt{3})-(\sqrt{10}-\sqrt{7})=(\sqrt{5}+\sqrt{7})-(\sqrt{3}+\sqrt{10})
Mà:
\((\sqrt{5}+\sqrt{7})^2=12+\sqrt{35}< 12+\sqrt{36}=18\)
\((\sqrt{3}+\sqrt{10})^2=13+\sqrt{30}>13+\sqrt{25}=18\)
\(\Rightarrow \sqrt{3}+\sqrt{10}> \sqrt{5}+\sqrt{7}\Rightarrow \sqrt{5}-\sqrt{3}< \sqrt{10}-\sqrt{7}\)
Lời giải:
a.
$5+\sqrt{2}>5+\sqrt{1}=6$
$4+\sqrt{3}< 4+\sqrt{4}=6$
$\Rightarrow 5+\sqrt{2}>4+\sqrt{3}$
b.
$\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$
$\sqrt{5}-\sqrt{3}=\frac{5-3}{\sqrt{5}+\sqrt{3}}=\frac{2}{\sqrt{5}+\sqrt{3}}< \frac{2}{\sqrt{2}}=\sqrt{2}$
Vậy $\sqrt{8}-\sqrt{2}>\sqrt{5}-\sqrt{2}$