f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)

mik sửa lại câu f , tí nhé :

f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

a,\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)

\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}\) \(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)

*\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}+1=2\)

\(\Rightarrow A\in Z\)

* \(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-2\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\) \(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\) \(=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)

\(=\dfrac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\) \(=\dfrac{3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}}{9-8}\)

\(=2\)

\(\Rightarrow B\in Z\)

A= căn (5-2 (căn 5) +1)-căn (5+2 (căn 5) +1)

=căn ((căn 5)-1)^2 -căn ((căn 5)+1)^2

=l (căn 5) -1l  -   l (căn 5) +1l

=căn 5 -1 -căn 5 -1 

=-2

A,  biến đổi 6= căn bậc hai của 5 + 1 -> hằng đẳng thức

Tính tiếp sẽ ra

a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

\(=3\sqrt{5}\)

b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

\(=3-2\sqrt{2}+2\sqrt{2}-1\)

=2

c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

\(=2-\sqrt{2}+3\sqrt{2}-2\)

\(=2\sqrt{2}\)

\(\sqrt{\left(2\sqrt{2-1}\right)^2}-\sqrt{17+12\sqrt{2}}\\ =\left|2\sqrt{2}-1\right|-\sqrt{9+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\\ =2\sqrt{2}-1-\sqrt{\left(3+2\sqrt{2}\right)^2}\\=2\sqrt{2}-1-\left(3+2\sqrt{2}\right)\\ =2\sqrt{2}-1-3-2\sqrt{2}\\ =-4\)

__

\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\\ =\left|2-\sqrt{5}\right|+\sqrt{9-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =2-\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}\\ =2-\sqrt{5}+3-\sqrt{5}\\ =5-2\sqrt{5}\)

__

\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\\ =\left|4-3\sqrt{2}\right|-\sqrt{18+2\cdot3\cdot\sqrt{2}+1}\\ =4-3\sqrt{2}-\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =4-3\sqrt{2}-3\sqrt{2}-1\\ =3-6\sqrt{2}\)

3 bài đầu dễ tự làm nhé.

Bài 4:

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

\(=\dfrac{\sqrt{\left(1-\sqrt{2}\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(1+\sqrt{2}\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)

\(=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{1+\sqrt{2}}{3+2\sqrt{2}}\)

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

\(=3\sqrt{2}+4-3-2\sqrt{2}-\left(3-2\sqrt{2}+3\sqrt{2}-4\right)\)

\(=3\sqrt{2}+4-3-2\sqrt{2}-\left(-1+\sqrt{2}\right)\)

\(=3\sqrt{2}+4-3-2\sqrt{2}+1-\sqrt{2}\)

\(=0+2\)

\(=2\)

Vậy B là số tự nhiên.

1.

a) nhân cả tử lẫn mẫu với 1+ \(\sqrt{2}-\sqrt{5}\)

b) tương tự a

2.

a) tách 29 = 20 + 9 là ra hằng đẳng thức, tiếp tục.

1.

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

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

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

b) \(\dfrac{1}{\sqrt{x}+\sqrt{x+1}}=\dfrac{\sqrt{x}-\sqrt{x+1}}{\left(\sqrt{x}+\sqrt{x+1}\right)\left(\sqrt{x}-\sqrt{x+1}\right)}\)

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

2.

a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)

=\(\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)=\(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

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

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

b)\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

=\(\sqrt{6+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)

=\(\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)

=\(\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\)

=\(\sqrt{6+2\sqrt{4-\sqrt{12}}}\)

=\(\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

=\(\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}\)

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

c) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)

làm giống câu a

3. a=\(\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

=\(\sqrt{3-\sqrt{5}}\left(3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\)

=\(\sqrt{3-\sqrt{5}}\left(2\sqrt{10}+2\sqrt{2}\right)\)

=\(\sqrt{3-\sqrt{5}}.\sqrt{2}\left(2\sqrt{5}+2\right)\)

=\(\sqrt{6-2\sqrt{5}}\left(2\sqrt{5}+2\right)=\left(\sqrt{5}-1\right)\left(2\sqrt{5}+2\right)\)

=\(10-2\sqrt{5}+2\sqrt{5}-2=8\)

vậy a là số tự nhiên

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

Vì \(\left\{{}\begin{matrix}2=\sqrt{4}< \sqrt{5}\\2\sqrt{2}=\sqrt{8}>\sqrt{5}\end{matrix}\right.\) nên A = \(\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)

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

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

b) B = \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\) (B > 0)

Ta có:

B² = \(6+2\sqrt{5}-2\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}+6-2\sqrt{5}\)

     = \(12-2\sqrt{36-20}\)

     = \(12-8\)

     = \(4\)

\(\Rightarrow\) B =\(\pm2\) nhưng vì B > 0 nên B = 2

Vậy B = 2

c) C = \(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\) (C > 0)

Ta có: 

C² = \(17+12\sqrt{2}+2\sqrt{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}+\left(17-12\sqrt{2}\right)\)

     = \(34+2\sqrt{289-288}\)

     = \(34+2\)

     = \(36\)

\(\Rightarrow C=\pm6\) nhưng vì C > 0 nên C = 6