Question

so sánh \(\sqrt{5}\) và \(\sqrt{6}-1\)

Answer 1

Giả sử : \(\sqrt{6}-1>\sqrt{5}\) ⇔ \(\left(\sqrt{6}-1\right)^2>1\) ⇔ \(6-2\sqrt{6}+1>1\)⇔ \(\sqrt{36}-\sqrt{24}>0\) ( Luôn đúng )

KL : Vậy : \(\sqrt{6}-1>\sqrt{5}\)

Answer 2

Giả sử: \(\sqrt{5}< \sqrt{6}-1\Rightarrow\left(\sqrt{5}\right)^2< \left(\sqrt{6}-1\right)^2\)

\(\Rightarrow7-2\sqrt{6}-5>0\Rightarrow2-2\sqrt{6}>0\) vô lí( vì \(\sqrt{4}< \sqrt{24}\))

Vậy \(\sqrt{5}>\sqrt{6}-1\)

Answer 3

Ta có \(\left(\sqrt{5}\right)^2=5\)

\(\left(\sqrt{6}-1\right)^2=6-2\sqrt{6}+1=7-2\sqrt{6}=5+2-2\sqrt{6}\)

Ta có 4<24⇒\(\sqrt{4}< \sqrt{24}\Rightarrow2< 2\sqrt{6}\Rightarrow2-2\sqrt{6}< 0\Rightarrow5+2-2\sqrt{6}< 5\Rightarrow6-2\sqrt{6}+1< 5\Rightarrow\sqrt{6-2\sqrt{6}-1}< \sqrt{5}\Rightarrow\sqrt{6}-1< \sqrt{5}\)

Answer 4

Giả sử √5 > √6 -1 ⇔ √5²>√6² ⇔ 5>6(BDT sai)

Vậy √5<√6