Question

so sánh giúp mk nha

\(\sqrt{27}+\sqrt{6}so\sqrt{48}\)

Answer 1

\(\sqrt{27+6}=\sqrt{33}\)

\(\sqrt{33}< \sqrt{48}\) 

 

Answer 2

27>25>0

→\(\sqrt{27}\)>\(\sqrt{25}\)

\(\sqrt{27}\)>5

6>4>0

\(\sqrt{6}\)>\(\sqrt{4}\) 

\(\sqrt{6}\)>2

\(\sqrt{27}\)+\(\sqrt{6}\)>2+5→\(\sqrt{27}\)+\(\sqrt{6}\)>7

0<48<49→\(\sqrt{48}\)<\(\sqrt{49}\)→\(\sqrt{48}\)<7

Từ đó suy ra \(\sqrt{27}\)+\(\sqrt{6}\)>\(\sqrt{48}\)