\(\sqrt{2}+\sqrt{3}+\sqrt{8}+\sqrt{11}< \sqrt{4}+\sqrt{4}+\sqrt{9}+\sqrt{16}=2+2+3+4=11\)
Ta có : \(\sqrt{2}+\sqrt{3}+2\sqrt{2}+\sqrt{23}< \sqrt{3}+\sqrt{3}+2\sqrt{3}+\sqrt{25}\)
\(=4\sqrt{3}+5< 4\sqrt{4}+5=4.2+5=6+5=11\)
Vậy \(\sqrt{2}+\sqrt{3}+\sqrt{8}+\sqrt{23}< 11\)
Cách khác
Giải:
Ta có:
\(\sqrt{2}< \sqrt{2,25}=1,5\)
\(\sqrt{3}< \sqrt{3,24}=1,8\)
\(\sqrt{8}< \sqrt{8,41}=2,9\)
\(\sqrt{23}< \sqrt{23,04}=4,8\)
Cộng theo vế, ta được:
\(\sqrt{2}+\sqrt{3}+\sqrt{8}+\sqrt{23}< 1,5+1,8+2,9+4,8\)
\(\Leftrightarrow\sqrt{2}+\sqrt{3}+\sqrt{8}+\sqrt{23}< 11\)
Vậy ...
