Question

chứng minh rằng ,với a>b>0 thì \(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)

Answer 1

Có: \(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\)

\(\left(\sqrt{a-b}\right)=a-b=a+b-2b\)

Vì: \(a>b>0\rightarrow ab>b^2\rightarrow2\sqrt{ab}>2\sqrt{b^2}\rightarrow2\sqrt{ab}>2b\)

\(\Rightarrow-2\sqrt{ab}< -2b\Rightarrow a-2\sqrt{ab}+b< a-2b+b\)

\(\Rightarrow\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)

Nhớ tick mik nha

Answer 2

Có \(a>b>0\)

<=> \(\sqrt{a}>\sqrt{b}>0\)

Với a>b>0 có: \(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\) (1)

<=> \(a+b-2\sqrt{ab}< a-b\)

<=> \(0< a-b-a-b+2\sqrt{ab}\)

<=> \(0< -2b+2\sqrt{ab}\)

<=> \(0< \sqrt{ab}-b\)

<=> \(0< \sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)(luôn đúng vì \(\sqrt{a}>\sqrt{b}>0\))

Vậy (1) đc CM