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

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

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

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

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

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

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

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

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

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

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

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

b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)

Bài 1: 

Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)

Số giá trị nguyên thỏa mãn điều kiện là:

\(\left(2+4\right)+1=7\)

Câu 4: a) ĐK: \(x^2\ge9\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

b) ĐK: \(x^2-3x+2\ge0\Leftrightarrow\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)

c) Đk: \(-3\le x< 5\)

d) x + 3 và 5 - x đồng dấu. Xét hai trường hợp:

\(\left\{{}\begin{matrix}x+3\ge0\\5-x>0\left(\text{do mẫu phải khác 0}\right)\end{matrix}\right.\Leftrightarrow-3\le x< 5\)

\(\left\{{}\begin{matrix}x+3< 0\\5-x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -3\\x>5\end{matrix}\right.\) do x ko thể đồng thời thỏa mãn cả hai nên loại.

Câu 1:

a) Đặt \(A=x+\sqrt{\left(x+2\right)^2}\cdot\left(x-2\right)\)

\(A=x+\left|x+2\right|\cdot\left(x-2\right)\)

+) Với \(x\ge-2\):

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

+) Với \(x< -2\):

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

b) \(B=\sqrt{m^2-6m+9-2m}\)

\(B=\sqrt{m^2-8m+9}\)

Bạn xem lại đề nhé :)

c) \(C=1+\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)

\(C=1+\sqrt{x-1}\)

d) \(D=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

\(D=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)

\(D=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)

\(D=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)

+) Xét \(x\ge8\):

\(D=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)

+) Xét \(4< x< 8\):

\(D=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)

Vậy....

Câu 2:

a) Ta có: \(\sqrt{5}< \sqrt{9}=3\)

b) \(2\sqrt{2}< \left(2+1\right)\sqrt{2}=3\sqrt{2}\)

c) \(-4\sqrt{5}>-4\sqrt{6}>-6\sqrt{6}\)

d) Xét hiệu: \(2\sqrt{3}-5-\sqrt{3}+4=\sqrt{3}-1>\sqrt{1}-1=0\)

Nên \(2\sqrt{3}-5>\sqrt{3}-4\)

e) Tương tự