Question

So sanh

16 va \(\sqrt{15}\). \(\sqrt{17}\)

Answer 1

Ta có :

√15.√17= √16-1.√16+1

=√16²-1

Vì 16²-1 < 16² nên

√16²-1< √16²

^{Vậy 16> √15.√17}

Answer 2

\(\sqrt{15}\cdot\sqrt{17}=\sqrt{255}< \sqrt{256}=16\)

Answer 3

\(16=\sqrt{16^2}\)

\(16^2=\left(15+1\right).\left(17-1\right)=15.17-15+17=15.17+2\)

Mà \(15.17+2>15.17\)

\(\Leftrightarrow\sqrt{15.17+2}>\sqrt{15.17}\)

\(\Leftrightarrow16>\sqrt{15}.\sqrt{17}\)

Answer 4

Ta thấy ;

\(\sqrt{15}.\sqrt{17}\)

= \(\sqrt{15.17}\)

= \(\sqrt{255}\)

So sánh \(16;\sqrt{255}\)

ta thấy :

\(16^2=256\)

\(\sqrt{255^2}=255\)

Do \(256>255\)

=> \(16>\sqrt{15}.\sqrt{17}\)