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

\(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

Mà \(\sqrt{2004}+\sqrt{2003}< \sqrt{2006}< \sqrt{2005}\)

\(\Rightarrow\dfrac{1}{\sqrt{2004}+\sqrt{2003}}>\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

\(\Rightarrow\sqrt{2004}-\sqrt{2003}>\sqrt{2006}-\sqrt{2005}\)

2) \(-x^2+4x-2\)

\(=-\left(x^2-4x+2\right)\)

\(=-\left(x^2-4x+4-2\right)\)

\(=-\left(x-2\right)^2+2\)

Ta có: \(-\left(x-2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-2\right)^2+2\le2\forall x\)

Dấu "=" xảy ra:

\(\Leftrightarrow-\left(x-2\right)^2+2=2\Leftrightarrow x=2\)

Vậy: GTLN của bt là 2 tại x=2

b) \(\sqrt{2x^2-3}\) (ĐK: \(x\ge\sqrt{\dfrac{3}{2}}\))

Mà: \(\sqrt{2x^2-3}\ge0\forall x\)

Dấu "=" xảy ra:

\(\sqrt{2x^2-3}=0\Leftrightarrow x=\sqrt{\dfrac{3}{2}}=\dfrac{3\sqrt{2}}{2}\)

Vậy GTNN của bt là 0 tại \(x=\dfrac{3\sqrt{2}}{2}\)

...

1:

b: \(4\sqrt{5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{75}\)

=>\(4\sqrt{5}>5\sqrt{3}\)

=>\(\sqrt{4\sqrt{5}}>\sqrt{5\sqrt{3}}\)

c: \(3-2\sqrt{5}-1+\sqrt{5}=2-\sqrt{5}< 0\)

=>\(3-2\sqrt{5}< 1-\sqrt{5}\)

d: \(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

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

\(\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\)

=>\(\dfrac{1}{\sqrt{2006}+\sqrt{2005}}< \dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)

=>\(\sqrt{2006}-\sqrt{2005}< \sqrt{2005}-\sqrt{2004}\)

e: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)

\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)

=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)

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

Giả sử: \(\sqrt{2005}-\sqrt{2004}\le\sqrt{2004}-\sqrt{2003}\)

\(\Leftrightarrow\sqrt{2005}+\sqrt{2003}\le2\sqrt{2004}\)

\(\Leftrightarrow\left(\sqrt{2005}+\sqrt{2003}\right)^2\le\left(2\sqrt{2004}\right)^2\)

\(\Leftrightarrow2005+2\sqrt{2005.2003}+2003\le4.2004\)

\(\Leftrightarrow4008+2\sqrt{\left(2004+1\right)\left(2004-1\right)}\le4008+4008\)

\(\Leftrightarrow2\sqrt{2004^2-1}\le4008\)

\(\Leftrightarrow\sqrt{2004^2-1}\le2004\)

\(\Leftrightarrow\sqrt{2004^2-1}\le\sqrt{2004^2}\)

Vậy giả sử đúng

\(\Rightarrow\sqrt{2005}-\sqrt{2004}\le\sqrt{2004}-\sqrt{2003}\)

dùng sai dấu rồi ạ :)) dùng dấu <  thay cho dấu  ≤  nhé

Ta có: +) 2005 - 2004 = 1 \(\Leftrightarrow\left(\sqrt{2005}-\sqrt{2004}\right)\left(\sqrt{2005}+\sqrt{2004}\right)=1\)

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

+) \(2004-2003=1\)

\(\Leftrightarrow\left(\sqrt{2004}-\sqrt{2003}\right)\left(\sqrt{2004}+\sqrt{2003}\right)=1\)

\(\Leftrightarrow\sqrt{2004}-\sqrt{2003}=\dfrac{1}{\sqrt{2004}+\sqrt{2003}}\) (2)

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

\(\Rightarrow\dfrac{1}{\sqrt{2005}+\sqrt{2004}}< \dfrac{1}{\sqrt{2004}+\sqrt{2003}}\) (3)

Từ (1), (2) và (3) \(\Rightarrow\sqrt{2005}-\sqrt{2004}< \sqrt{2004}-\sqrt{2003}\)

Ta có : \(\sqrt{2005}-\sqrt{2004}\) ; \(\sqrt{2004}-\sqrt{2003}\)

=> \(\sqrt{2005}>\sqrt{2004}>\sqrt{2003}\)

=> \(\sqrt{2005}-\sqrt{2004}\)> \(\sqrt{2004}-\sqrt{2003}\)

\(\sqrt{2005}-\sqrt{2004}=0.01116778328\)

\(\sqrt{2004}-\sqrt{2003}=0.01117057\)

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

Bài này ta dùng phương pháp trục căn thức ở mẫu 

Ta có: \(\frac{1}{a}=\frac{1}{\sqrt{2004}-\sqrt{2003}}=\frac{\sqrt{2004}+\sqrt{2003}}{\left(\sqrt{2004}-\sqrt{2003}\right)\left(\sqrt{2004}+\sqrt{2003}\right)}\)

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

Tương tự: 1/b = căn 2005 + căn 2004

Vì căn 2004 + căn 2003 < căn 2005 + căn 2004

=> căn 2004 - căn 2003 > căn 2005 - căn 2004

Vậy a > b

P/s: Bài giải còn nhiều sai sót, mong các anh chị thông cảm và sửa cho em.

Giải

a) Ta có: \(\left(\sqrt{a+b}\right)^2=a+b\) (1)

\(\left(\sqrt{a}+\sqrt{b}\right)^2=a+b+2\sqrt{ab}\) (2)

Vì a > 0, b > 0 nên \(2\sqrt{ab}>0\), do đó từ (1) và (2) suy ra

\(\left(\sqrt{a+b}\right)^2< \left(\sqrt{a}+\sqrt{b}\right)^2\) hay \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)

b) Áp dụng câu a) cho hai số dương 2004 và 2005, ta có

\(\sqrt{2004+2005}< \sqrt{2004}+\sqrt{2005}\)