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.

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

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

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

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

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

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

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

\(=2\cdot2004+2\sqrt{2004^2-1}\)

Mà \(2004^2-1< 2004^2\Rightarrow2\cdot2004+2\sqrt{2004^2-1}< 2\cdot2004+2\sqrt{2004^2}=2\cdot2004+2\cdot2004=4\cdot2004\)

Mặt khác \(\left(2\sqrt{2004}\right)^2=4\cdot2004\)

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

ta có

Áp dụng BĐT CAuchy-Schwarz ta có:

Đặt \(A^2=\left(\sqrt{2003}+\sqrt{2005}\right)^2\)

\(\le\left(1+1\right)\left(2003+2005\right)\)

\(=2\cdot4008=8016\)

\(\Rightarrow A^2\le8016\Rightarrow A\le2\sqrt{2004}=B\)

MÌNH LỚP 7 NHƯNG TRẢ LỜI ĐƯỢC LÈ

ÂY ... >>>>>>

BI ...========

CI <<<<<<<<<

CÂU TRẢ LỜI LÀ Â B C D E F J A T O E M S D

ÂYY

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

ta có :\(\left(\sqrt{2005}+\sqrt{2003}\right)^2\le\left(1^2+1^2\right)\left(2005+2003\right)=2.4008\)(bđt bu-nhia-cop xki)

\(\left(2\sqrt{2004}\right)^2=4.2004=2.4008\)

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

ta có: A=\(\left(\sqrt{2003}+\sqrt{2005}\right)^2=2003+2005+2\sqrt{2003.2005}\)

=\(4008+2\sqrt{2003.2005}\)

=4008+\(2\sqrt{2003.2004+2003}\)(1)

B=\(\left(2\sqrt{2004}\right)^2=8016=4008+2.2004\)

=\(4008+2\sqrt{2004^2}\)

=4008+\(2\sqrt{2003.2004+2004}\)(2)

từ (1) và (2) ==>A<B

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

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

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

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

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

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

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

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

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

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

2)

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

\(=4008+2\sqrt{2004^2}\)

Ta có \(2004^2>2004^2-1\Rightarrow\sqrt{2004^2}>\sqrt{2004^2-1}\Rightarrow4008+2\sqrt{2004^2}>4008+2\sqrt{2004^2-1}\)

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

1.  a) 108
     b) 128
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