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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 : \(x^2\ge x^2-1\Rightarrow x^2>\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow4008+2\sqrt{2004.2004}>4008+2\sqrt{2003.2005}\)
\(\Rightarrow2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}\)
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
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 ) \(\sqrt{2}+\sqrt{3}\) 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 \(\sqrt{5\sqrt{3}}>\sqrt{3\sqrt{5}}\)
Võ Đông Anh Tuấn
Áp dụng \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)
a)
\(7=\sqrt{49}\\ 3\sqrt{5}=\sqrt{9}\cdot\sqrt{5}=\sqrt{9\cdot5}=\sqrt{45}\\ \text{Vì }\sqrt{49}>\sqrt{45}\text{ nên }7>3\sqrt{5}\)
Vậy \(7>3\sqrt{5}\)
b)
\(2\sqrt{7}+3=\sqrt{4}\cdot\sqrt{7}+3=\sqrt{4\cdot7}+3=\sqrt{28}+3\\ \sqrt{28}+3>\sqrt{25}+3=5+3=8\)
Vậy \(8< 2\sqrt{7}+3\)
c)
\(3\sqrt{6}=\sqrt{9}\cdot\sqrt{6}=\sqrt{9\cdot6}=\sqrt{54}\\ 2\sqrt{15}=\sqrt{4}\cdot\sqrt{15}=\sqrt{4\cdot15}=\sqrt{60}\\ \text{Vì } \sqrt{54}< \sqrt{60}\text{nên }3\sqrt{6}< 2\sqrt{15}\)
Vậy \(3\sqrt{6}< 2\sqrt{15}\)
Áp dụng BĐT CAuchy-Schwarz ta có:
Đặt \(A^2=\left(\sqrt{2003}+\sqrt{2005}\right)^2\)
\(\le\left(1+1\right)\left(2003+2005\right)\)
\(=2\cdot4008=8016\)
\(\Rightarrow A^2\le8016\Rightarrow A\le2\sqrt{2004}=B\)
a: \(1-\sqrt{3}=\dfrac{-2}{1+\sqrt{3}}\)
\(2-\sqrt{6}=\dfrac{-2}{2+\sqrt{6}}\)
mà 1+căn 3<2+căn 6
nên 1-căn 3>2-căn 6
b: \(\left(2-\sqrt{2}\right)^2=6-4\sqrt{2}\)
(1/2)^2=1/4
mà 6-4căn 2-1/4>0
nên 2-căn 2>1/2