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

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

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

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

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

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}×\sqrt{2004-2\sqrt{2006}-2\sqrt{2005}}=\sqrt{2004-2\sqrt{2006-2\sqrt{2005}}}\)

lấy vế đầu trừ vế sau nếu kết quả dương suy ra vế đầu lớn hơn nếu kq âm thì vế sau lớn hơn

có\(\sqrt{2006}-\sqrt{2005}=\frac{\left(\sqrt{2006}-\sqrt{2005}\right)\left(\sqrt{2006}+\sqrt{2005}\right)}{\sqrt{2006}+\sqrt{2005}}\)\(=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)

có\(\sqrt{2005}-\sqrt{2004}=\frac{\left(\sqrt{2005}-\sqrt{2004}\right)\left(\sqrt{2005}+\sqrt{2004}\right)}{\sqrt{2005}+\sqrt{2004}}\)\(=\frac{1}{\sqrt{2005}+\sqrt{2004}}\)

ta lại có 2006>2005\(\Rightarrow\sqrt{2006}>\sqrt{2005}\)có 2005>2004\(\Rightarrow\sqrt{2005}>\sqrt{2004}\)

\(\Rightarrow\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\)\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}< \frac{1}{\sqrt{2005}+\sqrt{2004}}\)

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

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 : \(\sqrt{2004}+\sqrt{2006}< 2\sqrt{2005}\)

\(\Leftrightarrow2004+2006+2\sqrt{2004.2006}< 4.2005\)

\(\Leftrightarrow\sqrt{2004.2006}< 2005\Leftrightarrow2004.2006< 2005^2\)

\(\Leftrightarrow\left(2005-1\right)\left(2005+1\right)< 2005^2\)

\(\Leftrightarrow2005^2-1< 2005^2\) . BĐT đúng

Vậy \(\sqrt{2004}+\sqrt{2006}< 2\sqrt{2005}\)

Giả sử : \(\sqrt{2004}+\sqrt{2006}< 2\sqrt{2005}\) 

\(\Leftrightarrow2004+2006+2\sqrt{2004.2006}< 4.2005\)

\(\Leftrightarrow\sqrt{2004.2006}< 2005\Leftrightarrow2004.2006< 2005^2\)

\(\Leftrightarrow\left(2005-1\right)\left(2005+1\right)< 2005^2\)

\(\Leftrightarrow2005^2-1< 2005^2.\) BĐT đúng

Vậy \(\sqrt{2004}+\sqrt{2006}< 2\sqrt{2005}\)

a/ ĐKXĐ: ...

Đặt \(\sqrt{x+2006}=a\ge0\Rightarrow a^2-x=2006\)

Pt trở thành:

\(x^2+a=a^2-x\)

\(\Leftrightarrow x^2-a^2+x+a=0\)

\(\Leftrightarrow\left(x+a\right)\left(x-a+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-x\\a=x+1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2006}=-x\left(x\le0\right)\\\sqrt{x+2006}=x+1\left(x\ge-1\right)\end{matrix}\right.\) (1)

\(\Leftrightarrow\left[{}\begin{matrix}x+2006=x^2\\x+2006=\left(x+1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2006=0\\x^2+x-2005=0\end{matrix}\right.\)

Nhớ loại nghiệm của từng pt phù hợp với (1)

b/ ĐKXĐ: ...

Đặt \(\sqrt{1-\sqrt{x}}=a\Rightarrow\sqrt{x}=1-a^2\Rightarrow x=\left(1-a^2\right)^2\) (với \(0\le a\le1\))

\(\left(1-a^2\right)^2=\left(2005-a^2\right)\left(1-a\right)\)

\(\Leftrightarrow\left(1+a\right)^2\left(1-a\right)^2=\left(2005-a^2\right)\left(1-a\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\\left(1-a\right)\left(1+a\right)^2=2005-a^2\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow a^3-a+2004=0\)

Do \(0\le a\le1\Rightarrow a^3-a+2004>0\Rightarrow\) pt vô nghiệm

Vậy pt có nghiệm duy nhất \(x=0\)

@Nguyễn Việt Lâm

a) \(\sqrt{3x-4}\) + \(\sqrt{4x+1}\) = \(-16x^2 - 8x +1\) với

ĐKXĐ :

- Vế trái \(x \ge \frac{4}{3}\)

- Vế phải : \(-16x^2 - 8x +1\) \(\ge 0\) \(\Leftrightarrow \) \(x \le \frac{\sqrt{2}-1}{4}\) hoặc \(x \le \frac{-\sqrt{2}-1}{4}\)

Hai điều kiện trái ngược nhau

Vậy phương trình vô nghiệm .