Question

Chứng minh rằng:

\(\sqrt{2000}-2\sqrt{2001}+\sqrt{2002}< 0\)

Answer 1

Ta có:
\(\sqrt{2002}-\sqrt{2001}=\dfrac{1}{\sqrt{2002}+\sqrt{2001}}\)

\(\sqrt{2001}-\sqrt{2000}=\dfrac{1}{\sqrt{2001}+\sqrt{2000}}\)

Do \(\sqrt{2002}+\sqrt{2001}>\sqrt{2001}+\sqrt{2000}\)

\(\Rightarrow\dfrac{1}{\sqrt{2002}+\sqrt{2001}}< \dfrac{1}{\sqrt{2001}+\sqrt{2000}}\)

hay \(\sqrt{2002}-\sqrt{2001}\) < \(\sqrt{2001}-\sqrt{2000}\)

\(\Rightarrow\sqrt{2002}-2\sqrt{2001}+\sqrt{2000}< 0\) (đpcm)