a: 6>căn 5
=>6+2>2+căn 5
=>8>2+căn 5
b: căn 2>1
=>1+căn 2>2
a) Ta có: \(8=6+2\)
Do: \(6>5\Leftrightarrow6>\sqrt{5}\)
\(\Leftrightarrow6+2>\sqrt{5}+2\)
\(\Leftrightarrow8>2+\sqrt{5}\)
b) Ta có: \(2=1+1=1+\sqrt{1}\)
Do: \(1< 2\Leftrightarrow\sqrt{1}< \sqrt{2}\)
\(\Leftrightarrow1< \sqrt{2}\Leftrightarrow1+1< 1+\sqrt{2}\)
\(\Leftrightarrow2< 1+\sqrt{2}\)