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

\(\Rightarrow A^2=6+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}=6+2\sqrt{4-2\sqrt{3}}\)

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

\(\Rightarrow A^2-2A-2=4+2\sqrt{3}-2\left(1+\sqrt{3}\right)-2=0\)

thanks

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

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

\(\Rightarrow a^2=3+\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}+2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}\)\(\Rightarrow VT=3+\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}+2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}-2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}-2\)

\(=6-2=4\) ??? đề bài có sai ko bn?

\(a^2=6+2\sqrt{9-\left(5+2\sqrt{3}\right)}=6+2\sqrt{4-2\sqrt{3}}=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)

\(\Rightarrow a=\sqrt{3}+1\)

\(\Rightarrow a^2-2a-2=\left(a-1\right)^2-3=\left(\sqrt{3}+1-1\right)^2-3=3-3=0\)

Ta có a² = 6 + 2\(\sqrt{4-2\sqrt{3}}\)= 6 + \(2\sqrt{3}\)- 2 = 4 + 2\(\sqrt{3}\)= (\(\sqrt{3}\)+ 1)²

=> a = \(1+\sqrt{3}\)

Từ đó => a2​- 2a - 2 = 0

Cái đề bạn bị sai rồi nhé

Lời giải:

Ta có:

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

$=6+2\sqrt{3^2-(5+2\sqrt{3})}=6+2\sqrt{4-2\sqrt{3}}=6+2\sqrt{3+1-2\sqrt{3}}$

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

$\Rightarrow a=\sqrt{3}+1$ (do $a\geq 0$)

Do đó:

$a^2-2a-2=4+2\sqrt{3}-2(\sqrt{3}+1)-2=0$ (đpcm)

B xem lại đề bài thử nhé

bài này mình cũng dò lại đề rồi mình chép đúng đấy mà không làm được nên mới nhờ giải

Cố gắng hơn nữa bạn cho mình biết là cái đề này bạn chép từ bộ đề nào để mình lên mạng tìm thử xem sao. Biết đâu cái đề bạn cầm trên tay nó bị lỗi đánh máy thì sao.

\(\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}\)

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

\(=\dfrac{3-\sqrt{5}}{\sqrt{5}-3}\)

= - 1

\(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)

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

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

\(\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}\)

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

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

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

= 2

\(\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}\)

\(=4+9+16+49\)

= 78

\(\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y}\)

\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+\sqrt{xy}+y}\)

\(=\sqrt{x}-\sqrt{y}\)

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

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

\(\left[-\text{tử}-\right]=\sqrt{2}\left(2+\sqrt{3}\right)-\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^2}+\sqrt{2}\left(2-\sqrt{3}\right)+\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)^2}\)

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

\(\left[-\text{mẫu}-\right]=2-\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}-\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)

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

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

= 3

Ta có:

\(\dfrac{4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{3}\)

\(=\dfrac{8-\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{3\sqrt{2}}\)

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

\(=\dfrac{8-\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)}{3\sqrt{2}}=\dfrac{6}{3\sqrt{2}}=\sqrt{2}\)

\(\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}\)

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

\(=\dfrac{\left|\sqrt{a}-\sqrt{2}\right|}{\left|\sqrt{a}-\sqrt{3}\right|}\)