Question

so sánh : a/ \(\sqrt{3}\) +\(\sqrt{5}\) và \(\sqrt{17}\) ;b/ \(\sqrt{1999}\) + \(\sqrt{2001}\) và \(2\sqrt{200}\)  ;c/ \(\sqrt{2004}\)+ \(\sqrt{2006}\) và \(2\sqrt{2005}\) ; d/ \(\sqrt{5}+2\) và \(\sqrt{3}+\sqrt{6}\)

Answer 1

a: \(\left(\sqrt{3}+\sqrt{5}\right)^2=8+\sqrt{60}\)

\(\left(\sqrt{17}\right)^2=17=8+\sqrt{81}\)

mà 60<81

nên \(3+\sqrt{5}< \sqrt{17}\)

c: \(\left(\sqrt{2004}+\sqrt{2006}\right)^2=4010+2\cdot\sqrt{2005^2-1}\)

\(\left(2\cdot\sqrt{2005}\right)^2=8020=4010+2\cdot\sqrt{2005^2}\)

mà \(2005^2-1< 2005^2\)

nên \(\sqrt{2004}+\sqrt{2006}< 2\sqrt{2005}\)

d: \(\left(\sqrt{5}+2\right)^2=9+4\sqrt{5}=9+\sqrt{80}\)

\(\left(\sqrt{3}+\sqrt{6}\right)^2=9+2\cdot\sqrt{3\cdot6}=9+\sqrt{72}\)

mà 80>72

nên \(\sqrt{5}+2>\sqrt{3}+\sqrt{6}\)