Đặt A = \(\sqrt{ }\)2003 + \(\sqrt{ }\)2005 ; B = 2\(\sqrt{ }\)2004
A² = 2003 + 2005 + 2\(\sqrt{ }\)(2003.2005)
= 4008 + 2\(\sqrt{ }\)[(2004-1)(2004+1)]
= 4008 + 2\(\sqrt{ }\)(2004² - 1) < 2.2004 + 2\(\sqrt{ }\)(2004²) = 4.2004 = B²
\(\Rightarrow\) A < B

Ta có: \(2\sqrt{2003.2005}=2\sqrt{2004^2-1}< 2\sqrt{2004^2}\)

\(\Rightarrow\) 2003 + \(2\sqrt{2003.2005}+2005\) < 2003 + 4008 + 2005

hay \(\left(\sqrt{2003}+\sqrt{2005}\right)^2< 8016\)

\(\Rightarrow\) \(\sqrt{2003}+\sqrt{2005}\) < 2 \(\sqrt{2004}\)

Ta có

\(\sqrt{2005}-\sqrt{2004}=\dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)

và \(\sqrt{2004}-\sqrt{2003}=\dfrac{1}{\sqrt{2004+\sqrt{2003}}}\)

Quy về so sánh

\(\dfrac{1}{\sqrt{2005}+\sqrt{2004}}\) với \(\dfrac{1}{\sqrt{2004}+\sqrt{2003}}\)

Khi đó ,ta thấy ngay ở biểu thức thứ nhất lớn hơn mẫu ở biểu thức thứ hai ,các số này đều dương nên suy ra

\(\sqrt{2005}-\sqrt{2004}< \sqrt{2004}-\sqrt{2003}\)

Đặt A = \(\sqrt{2003}+\sqrt{2005}\)

B = \(2\sqrt{2004}\)

\(\Rightarrow A^2=2003+2005+2\sqrt{\left(2003.2005\right)}=4008+2\sqrt{\left(2004-1\right)\left(2004+1\right)}\)

\(=4008+2\sqrt{\left(2004^2-1\right)}\)

\(\Rightarrow B^2=4.2004=2.2004+2.2004=4008+2\sqrt{2004^2}\)

mà \(\sqrt{2004^2>\sqrt{ }2004^2-1}\)

\(\Rightarrow B^2>A^2\Rightarrow B>A\Rightarrow2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}\)

Nhớ k cho mình nhé! Thank you!!!

cần lắm một người nào đó giúp mình,hạn chót là ngày mai rồi

a) \(9=6+3=6+\sqrt{9}\)

\(6+2\sqrt{2}=6+\sqrt{8}\)

\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)

b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)

\(3^2=9=5+4=5+\sqrt{16}\)

\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)

c) \(9+4\sqrt{5}=\left(2+\sqrt{5}\right)^2\)

\(16=\left(2+2\right)^2=\left(2+\sqrt{4}\right)^2\)

\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)

\(2^2=14-10=14-\sqrt{100}\)

\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)

\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)

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 \(\sqrt{5\sqrt{3}}>\sqrt{3\sqrt{5}}\)

a) \(2\sqrt[3]{3}=\sqrt[3]{2^3}.\sqrt[3]{3}=\sqrt[3]{2^3.3}=\sqrt[3]{24}\)

Ta có : \(24>23\), nên \(\sqrt[3]{24}>\sqrt[3]{23}\)

Vậy \(2\sqrt[3]{3}>\sqrt[3]{23}\)

b) Ta có :

\(11=\sqrt[3]{11^3}=\sqrt[3]{1331}\)

Từ đó suy ra \(33< 3\sqrt[3]{1333}\)

Có:\(\sqrt{2005}-\sqrt{2004}=\frac{2005-2004}{\sqrt{2005}+\sqrt{2004}}=\frac{1}{\sqrt{2005}+\sqrt{2004}}\)

;\(\sqrt{2004}-\sqrt{2003}=\frac{2004-2003}{\sqrt{2004}+\sqrt{2003}}=\frac{1}{\sqrt{2004}+\sqrt{2003}}\)

\(\sqrt{2005}+\sqrt{2004}>\sqrt{2004}+\sqrt{2003}\)\(\Rightarrow\sqrt{2005}-\sqrt{2004}< \sqrt{2004}-\sqrt{2003}\)