a) \(\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)
Vậy \(\sqrt{7}+\sqrt{15}< 7\)
b) \(\sqrt{17}+\sqrt{5}+1>\sqrt{16}+\sqrt{4}+1=4+2+1=7\)\(=\sqrt{49}>\sqrt{45}\)
c) \(\dfrac{23-2\sqrt{19}}{3}< \dfrac{23-2\sqrt{16}}{3}=\dfrac{23-2.4}{3}=5=\sqrt{25}< \sqrt{27}\)
d) Giả sử \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\Leftrightarrow\left(\sqrt{3\sqrt{2}}\right)^2>\left(\sqrt{2\sqrt{3}}\right)^2\)
\(\Leftrightarrow3\sqrt{2}>2\sqrt{3}\Leftrightarrow\sqrt{18}>\sqrt{12}\Leftrightarrow18>12\)
BĐT cuối cùng đúng nên \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
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