\(\sqrt{17-\sqrt{33}}\cdot\sqrt{17+\sqrt{33}}\)

\(=\sqrt{\left(17-\sqrt{33}\right)\left(17+\sqrt{33}\right)}\)

\(=\sqrt{17^2-\left(\sqrt{33}\right)^2}\)

\(=\sqrt{289-33}\)

\(=\sqrt{256}\)

\(=\sqrt{16^2}\)

\(=16\)

\(\sqrt{17-\sqrt{33}}\sqrt{17+\sqrt{33}}=\sqrt{\left(17-\sqrt{33}\right)\left(17+\sqrt{33}\right)}\)

\(=\sqrt{17^2-33}=\sqrt{256}=16\)

Giải:

\(\sqrt{42-10\sqrt{17}}+\sqrt{33-8\sqrt{17}}\)

\(=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{\left(4-\sqrt{17}\right)^2}\)

\(=\left|5-\sqrt{17}\right|+\left|4-\sqrt{17}\right|\)

\(=5-\sqrt{17}+\sqrt{17}-4\)

\(=1\)

Vậy ...

\(\sqrt{42-10\sqrt{17}}+\sqrt{33-8\sqrt{17}}=\sqrt{25-2.5.\sqrt{17}+17}+\sqrt{16-2.4.\sqrt{17}+17}=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{\left(4-\sqrt{17}\right)^2}=\left|5-\sqrt{17}\right|+\left|4-\sqrt{17}\right|=5-\sqrt{17}+\sqrt{17}-4=1\)

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:

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

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 \(4\sqrt{5}>5\sqrt{3}\)

1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)

 

Cách 1 :\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{\sqrt{5}^2-2\sqrt{5}+\sqrt{1}^2}-\sqrt{\sqrt{5}^2+2\sqrt{5}+\sqrt{1}^2}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{1}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\)

\(=|\sqrt{5}-\sqrt{1}|-|\sqrt{5}+\sqrt{1}|=\sqrt{5}-\sqrt{1}-\sqrt{5}-\sqrt{1}=-2\)

Cách 2 \(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)

\(< =>A^2=6-2\sqrt{5}-6-2\sqrt{5}+2\sqrt{36-20}\)

\(< =>A^2=8-2\sqrt{5}-2\sqrt{5}=8-2\left(2\sqrt{5}\right)=8-4\sqrt{5}\)

<=>...

\(B=\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)

\(=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{2}+\sqrt{1}}{\sqrt{17+12\sqrt{2}}}\)

\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\sqrt{17+12\sqrt{2}}-\left(\sqrt{2}+1\right)\sqrt{17-12\sqrt{2}}}{\sqrt{17^2-\left(12\sqrt{2}\right)^2}}\)

tự làm tiếp đi , mình lười viết

Có :

+) \(\sqrt{33}< \sqrt{36}\)

+) \(\sqrt{17}>\sqrt{15}\Rightarrow-\sqrt{17}< -\sqrt{15}\)

Cộng theo vế 2 bất pt :

\(\sqrt{33}-\sqrt{17}< \sqrt{36}-\sqrt{15}=6-\sqrt{15}\)

Vậy...

Có :

\(3\sqrt{2}=\sqrt{18}\)

\(2\sqrt{3}=\sqrt{12}\)

Mà \(\sqrt{18}>\sqrt{12}\Rightarrow3\sqrt{2}>2\sqrt{3}\)

\(\Rightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)