Ta có: \(X=\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)

<=> \(X^2=6-3\sqrt{2+\sqrt{3}}+2+\sqrt{2+\sqrt{3}}-2\sqrt{3}.\sqrt{4-\left(2+\sqrt{3}\right)}\)

<= \(X^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3}.\sqrt{2-\sqrt{3}}\)

<=> \(X^2=8-\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{6}\left(\sqrt{3}-1\right)\)

<=> \(X^2=8-4\sqrt{2}\)

<=> \(X^2-8=-4\sqrt{2}\)

=> \(X^4-16X+64=32\)

<=> \(X^4-16X^2+32=0\)

Vậy X là nghiệm phương trình \(X^4-16X^2+32=0\)

a) \(\left(\sqrt{\dfrac{9}{20}}-\sqrt{\dfrac{1}{2}}\right).\sqrt{2}=\sqrt{\dfrac{9}{20}.2}-\sqrt{\dfrac{1}{2}.2}=\sqrt{\dfrac{9}{10}}-1=\dfrac{3}{\sqrt{10}}-1\)

\(=\dfrac{3\sqrt{10}}{10}-1\)

b) \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)\sqrt{3}=\sqrt{12.3}+\sqrt{27.3}-\sqrt{3.3}\)

\(=\sqrt{36}+\sqrt{81}-\sqrt{9}=6+9-3=12\)

c) \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right)\sqrt{6}=\sqrt{\dfrac{8}{3}.6}-\sqrt{24.6}+\sqrt{\dfrac{50}{3}.6}\)

\(=\sqrt{16}-\sqrt{144}+\sqrt{100}=4-12+10=2\)

1, \(sin\left(x+\dfrac{\pi}{6}\right)+cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{6}}{2}\)

⇔  \(\dfrac{\sqrt{2}}{2}sin\left(x+\dfrac{\pi}{6}\right)+\dfrac{\sqrt{2}}{2}cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)

⇔ \(sin\left(x+\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{4}\)

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

⇔ \(\dfrac{\left(\sqrt{3}-1\right)}{2\sqrt{2}}sinx+\dfrac{\left(\sqrt{3}+1\right)}{2\sqrt{2}}cosx=\dfrac{1-\sqrt{3}}{2\sqrt{2}}\)

⇔ sinx . si