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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 \(a>0\) )
\(\Rightarrow a^2-2a-2=4+2\sqrt{3}-2\left(\sqrt{3}+1\right)-2=0\)
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}.}\)
\(\Rightarrow A^2=6+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}=6+2\sqrt{4-2\sqrt{3}}\)
\(\Leftrightarrow A^2=6+2\left(\sqrt{3}-1\right)=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow A=\sqrt{3}+1\)
\(\Rightarrow A^2-2A-2=4+2\sqrt{3}-2\left(1+\sqrt{3}\right)-2=0\)
\(a=\sqrt{3+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}+\sqrt{3-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}\)
\(a=\sqrt{3+\sqrt{3}+\sqrt{2}}+\sqrt{3-\sqrt{3}-\sqrt{2}}\)
\(\Rightarrow a^2=3+\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}+2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}\)\(\Rightarrow VT=3+\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}+2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}-2\sqrt{\left(3+\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{3}-\sqrt{2}\right)}-2\)
\(=6-2=4\) ??? đề bài có sai ko bn?
\(a^2=6+2\sqrt{9-\left(5+2\sqrt{3}\right)}=6+2\sqrt{4-2\sqrt{3}}=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)
\(\Rightarrow a=\sqrt{3}+1\)
\(\Rightarrow a^2-2a-2=\left(a-1\right)^2-3=\left(\sqrt{3}+1-1\right)^2-3=3-3=0\)
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)
a) \(\sqrt{\left(\sqrt{7-2}\right)^2}=\sqrt{5}\)
b)\(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-3\sqrt{2}\right)^2}\)
=\(\sqrt{2}-1-2+3\sqrt{2}=4\sqrt{2}-3\)
c)\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
=\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=2\sqrt{3}\)
d) hình như bn ghi sai
e)\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4-2\sqrt{3}}}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{4+2\sqrt{3}}}\right):\sqrt{2}\)
=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{3}+1}\right):\sqrt{2}\)
=\(\dfrac{\sqrt{2+\sqrt{3}}\left(\sqrt{3}+1\right)+\sqrt{2-\sqrt{3}}\left(\sqrt{3}-1\right)}{2\sqrt{2}}\)
=\(\dfrac{\sqrt{6+3}+\sqrt{2+\sqrt{3}}+\sqrt{6-3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)
=\(\dfrac{3+\sqrt{2+\sqrt{3}}+\sqrt{3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)
=\(\dfrac{3+\sqrt{3}}{2\sqrt{2}}\)
f) \(\sqrt{9a^2}+3a-7=-3a+3a-7=-7\)
g)\(\dfrac{\sqrt{4x^2-4x+1}}{4x-2}+3x+2\)
=\(\dfrac{\sqrt{\left(2x-1\right)^2}}{4x-2}+3x+2=\dfrac{2x-1}{2\left(2x-1\right)}+3x+2\)
=\(\dfrac{1}{2}+3x+2=\dfrac{5}{2}+3x\)
h)\(\sqrt{\left(5a-1\right)^2}+2a-3\)
nếu a<0 :\(-5a+1+2a-3=-3a-2\)
nếu a>0 : \(5a-1+2a-3=7a-4\)
i)\(\sqrt{\dfrac{2a}{5}}.\sqrt{\dfrac{5a}{18}}+2\left(a-1\right)\)
=\(\sqrt{\dfrac{10a^2}{90}}+2a-2=\sqrt{\dfrac{a^2}{9}}+2a-2\)
=\(\dfrac{a}{3}+2a-2=\dfrac{7a}{3}-2\)
Bình phương a ta được
\(a^2=3+3+\sqrt{5+2\sqrt{3}}-\sqrt{5+2\sqrt{3}}+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}\)
\(a^2=6+2\sqrt{9-3\sqrt{5+2\sqrt{3}}+3\sqrt{5+2\sqrt{3}}-5-2\sqrt{3}}\)
\(a^2=6+2\sqrt{9-5-2\sqrt{3}}\Rightarrow a^2=6+2\sqrt{4-2\sqrt{3}}\Rightarrow a^2=6+2\sqrt{3+1-2.1.\sqrt{3}}\)\(a^2=6+2\sqrt{\left(\sqrt{3}-1\right)^2}\Rightarrow a^2=6+2\sqrt{3}-2=4+2\sqrt{3}=3+1+2.1.\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow a=\sqrt{3}+1\)
Rồi bạn tự thay vào tính típ nha
Chúc bạn học tốt
T I C K ủng hộ nha
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