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a ) \(\sqrt{2}+\sqrt{3}\) và \(\sqrt{10}\)
Ta có : \(\left(\sqrt{2}+\sqrt{3}\right)^2=2+3+2\sqrt{6}=5+2\sqrt{6}\)\(=5+\sqrt{24}\)
\(\left(\sqrt{10}\right)^2=10=5+5=5+\sqrt{25}\)
Vì \(\sqrt{24}< \sqrt{25}\Rightarrow5+\sqrt{24}< 5+\sqrt{25}\)hay \(\sqrt{2}+\sqrt{3}< \sqrt{10}\)
b ) \(\sqrt{2003}+\sqrt{2005}\) và \(2\sqrt{2004}\)
Ta có : \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=2003+2005+2\sqrt{2003.2005}\)
\(=4008+2\sqrt{\left(2004-1\right)\left(2004+1\right)}\)
\(=4008+2\sqrt{2004^2-1}\)
\(\left(2\sqrt{2004}\right)^2=4.2004=2.2004+2\sqrt{2004^2}\)\(=4008+2\sqrt{2004^2}\)
Vì \(4008+2\sqrt{2004^2-1}< 4008+2\sqrt{2004^2}\)=> \(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
c ) \(\sqrt{5\sqrt{3}}\)và \(\sqrt{3\sqrt{5}}\)
Ta có : \(\sqrt{5\sqrt{3}}=\sqrt{\sqrt{5^2.3}}=\sqrt{\sqrt{75}}\)
\(\sqrt{3\sqrt{5}}=\sqrt{\sqrt{3^2.5}}=\sqrt{\sqrt{45}}\)
Vì 75 > 45 => \(\sqrt{75}>\sqrt{45}\)hay \(\sqrt{5\sqrt{3}}>\sqrt{3\sqrt{5}}\)
Áp dụng BĐT CAuchy-Schwarz ta có:
Đặt \(A^2=\left(\sqrt{2003}+\sqrt{2005}\right)^2\)
\(\le\left(1+1\right)\left(2003+2005\right)\)
\(=2\cdot4008=8016\)
\(\Rightarrow A^2\le8016\Rightarrow A\le2\sqrt{2004}=B\)
Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) (bạn tự c/m) với a = 2003 , b = 2005
được : \(\frac{\sqrt{2003}+\sqrt{2005}}{2}< \sqrt{\frac{2003+2005}{2}}\)
\(\Rightarrow\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
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 \(\sqrt{5}+2>\sqrt{3}+\sqrt{6}\)
a) Ta có :\(\left(\sqrt{2}+\sqrt{3}\right)^2=2+3+2\sqrt{2}\cdot\sqrt{3}=5+2\sqrt{6}>5=\left(\sqrt{5}\right)^2\)
\(\Rightarrow\left(\sqrt{2}+\sqrt{3}\right)^2>\left(\sqrt{5}\right)^2\Leftrightarrow\sqrt{2}+\sqrt{3}>\sqrt{5}\)
a) \(\sqrt{2}+\sqrt{3}>\sqrt{5}\)
b) \(\sqrt{2003}+\sqrt{2005}< 2.\sqrt{2004}\)
HOK TOT