Question

So sánh:

a) \(1-\sqrt{3}\) và \(2-\sqrt{6}\)

b) \(2-\sqrt{2}\) và \(\dfrac{1}{2}\)

c) \(\sqrt{2003}+\sqrt{2005}\) và \(2\sqrt{2004}\)

Answer 1

a) ta có \(2-\sqrt{6}=\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)>1-\sqrt{3}\)

b) ta có : \(2-\sqrt{2}=\dfrac{1}{2}\left(4-2\sqrt{2}\right)\)

mà ta có : \(2\sqrt{2}< 3\) (vì \(8< 9\))

\(\Rightarrow4-2\sqrt{2}>4-3>1\) \(\Rightarrow2-\sqrt{2}>\dfrac{1}{2}\)

c) ta có : \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\sqrt{2003.2005}\)

\(\left(2\sqrt{2004}\right)^2=8016=4008+4008=4008+2\sqrt{2004.2004}\)

mà ta có : \(x^2\ge x^2-1\Rightarrow x^2>\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow4008+2\sqrt{2004.2004}>4008+2\sqrt{2003.2005}\)

\(\Rightarrow2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}\)

Answer 2

Mysterious Person giúp mk