Cho \(a=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}\)
cm: \(a^2-2a-2=0\)
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WR
14 tháng 8 2017
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}.}\)
\(\Rightarrow A^2=6+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}=6+2\sqrt{4-2\sqrt{3}}\)
\(\Leftrightarrow A^2=6+2\left(\sqrt{3}-1\right)=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow A=\sqrt{3}+1\)
\(\Rightarrow A^2-2A-2=4+2\sqrt{3}-2\left(1+\sqrt{3}\right)-2=0\)
HT
16 tháng 6 2019
\(a=\sqrt{3+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}+\sqrt{3-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}\)
\(a=\sqrt{3+\sqrt{3}+\sqrt{2}}+\sqrt{3-\sqrt{3}-\sqrt{2}}\)
\(\Rightarrow a^2=3+\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}+2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}\)\(\Rightarrow VT=3+\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}+2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}-2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}-2\)
\(=6-2=4\) ??? đề bài có sai ko bn?
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 6 2019
\(a^2=6+2\sqrt{9-\left(5+2\sqrt{3}\right)}=6+2\sqrt{4-2\sqrt{3}}=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)
\(\Rightarrow a=\sqrt{3}+1\)
\(\Rightarrow a^2-2a-2=\left(a-1\right)^2-3=\left(\sqrt{3}+1-1\right)^2-3=3-3=0\)
9 tháng 9 2016
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
AH
Akai Haruma
Giáo viên
27 tháng 7 2020
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)
12 tháng 6 2017
bài này mình cũng dò lại đề rồi mình chép đúng đấy mà không làm được nên mới nhờ giải
PA
7 tháng 8 2017
\(\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{5}-3}\)
\(=\dfrac{3-\sqrt{5}}{\sqrt{5}-3}\)
= - 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
PA
7 tháng 8 2017
\(\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y}\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+\sqrt{xy}+y}\)
\(=\sqrt{x}-\sqrt{y}\)
\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)+\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(\left[-\text{tử}-\right]=\sqrt{2}\left(2+\sqrt{3}\right)-\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^2}+\sqrt{2}\left(2-\sqrt{3}\right)+\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)^2}\)
\(=4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(\left[-\text{mẫu}-\right]=2-\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}-\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(=2-\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-3}\)
\(=2-\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)-1\)
= 3
Ta có:
\(\dfrac{4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{3}\)
\(=\dfrac{8-\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{3\sqrt{2}}\)
\(=\dfrac{8-\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{3\sqrt{2}}\)
\(=\dfrac{8-\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)}{3\sqrt{2}}=\dfrac{6}{3\sqrt{2}}=\sqrt{2}\)
\(\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}\)
\(=\sqrt{\dfrac{\left(\sqrt{a}-\sqrt{2}\right)^2}{\left(\sqrt{a}-\sqrt{3}\right)^2}}\)
\(=\dfrac{\left|\sqrt{a}-\sqrt{2}\right|}{\left|\sqrt{a}-\sqrt{3}\right|}\)
ta có : \(a=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}\)
\(\Rightarrow a^2=6+2\sqrt{4-2\sqrt{3}}=6+2\sqrt{\left(\sqrt{3}-1\right)^2}=4+2\sqrt{3}\)
\(\Rightarrow a=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\) (do \(a>0\) )
\(\Rightarrow a^2-2a-2=4+2\sqrt{3}-2\left(\sqrt{3}+1\right)-2=0\)
Mysterious Person giúp mk nha