a/ Có \(2=5-3=\sqrt{25}-3\)
Mà \(\sqrt{25}>\sqrt{5}\Rightarrow2>\sqrt{5}-3\)
b/ \(2\sqrt{31}=\sqrt{124}\)
\(10=\sqrt{100}\)
\(\sqrt{124}>\sqrt{100}\Rightarrow2\sqrt{31}>10\)
c/ \(-3\sqrt{11}=-\sqrt{99}\)
\(-12=-\sqrt{144}\)
Có \(-\sqrt{99}>-\sqrt{144}\Rightarrow-3\sqrt{11}>-12\)
d/ \(2=1+1\)
Có \(\sqrt{2}>1;\sqrt{3}>1\Rightarrow\sqrt{2}+\sqrt{3}>2\)
đ/ \(7=4+3=\sqrt{16}+\sqrt{9}\)
Có \(\sqrt{7}< \sqrt{9};\sqrt{15}< \sqrt{16}\Rightarrow\sqrt{7}+\sqrt{15}< 7\)
e/ \(\sqrt{3}+4=\sqrt{3}+\sqrt{16}\)
Có \(\sqrt{3}>\sqrt{2};\sqrt{16}>\sqrt{11}\Rightarrow\sqrt{2}+\sqrt{11}< \sqrt{3}+4\)
a, Ta có:
\(\sqrt{5}< 5\)
\(\Leftrightarrow\sqrt{5}-3< 5-3\)
\(\Leftrightarrow\sqrt{5}-3< 2\)
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b,
\(\sqrt{31}< \sqrt{25}\)
\(\Leftrightarrow\sqrt{31}< 5\)
\(\Leftrightarrow2\sqrt{31}< 2.5\)
\(\Leftrightarrow2\sqrt{31}< 10\)
Vậy...