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

\(=2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}=2-\sqrt{3}+\sqrt{3}-1\left(\text{vì }\sqrt{3}>1\right)\)

\(=1\)

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{\left(2\sqrt{6}\right)^2-2.2\sqrt{6}.3+9}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3\left(\text{vì }3>\sqrt{6};2\sqrt{6}>3\right)\)

\(=\sqrt{6}\)

a/

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

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

\(=1\)

b/

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}\)

\(=3-\sqrt{6}+\sqrt{24}-3\)

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

tick cho mình nha

a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)

\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)

\(=33\sqrt{3}\cdot\sqrt{3}\)

=99

b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)

\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)

c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=36-36\sqrt{2}+18\sqrt{3}\)

d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)

\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)

a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)

   \(=28.3+9.3-4.3=99\)

b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)

  \(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)

d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

        \(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

        \(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)

        \(=19\sqrt{3\sqrt{2}}\)

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

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

\(=2\sqrt{2}\)

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

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

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

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

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

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

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

b) \(\sqrt{12-6\sqrt{3}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{9}-\sqrt{3}\right)^2}\)

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

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

\(=\sqrt{6-4\sqrt{2}+\sqrt{\left(\sqrt{18}-2\right)^2}}\)

\(=\sqrt{6-4\sqrt{2}+\sqrt{18}-2}\)

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

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

\(A=\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4+2\sqrt{3}}\)

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

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

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

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

\(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\left(\dfrac{\sqrt{3}-1}{4}\right)^2}=\dfrac{\sqrt{12+2\sqrt{3}}}{4}\)

\(\Rightarrow2\cos\alpha=\dfrac{\sqrt{12+2\sqrt{3}}}{2}\). Chọn B.

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

\(=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}\cdot1+1^2}+\left|\sqrt{2}-2\right|\)

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

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

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

\(=3\)

b) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)

\(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{\dfrac{30}{15}}+\sqrt{\dfrac{144}{6}}\)

\(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}\)

\(=-8\sqrt{6}+2\sqrt{6}\)

\(=-6\sqrt{6}\)

c) \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)

\(=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}-2\right]\left[\dfrac{4\left(1-\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+4\right]\)

\(=\left(\sqrt{5}-1-2\right)\left(\dfrac{4\left(1-\sqrt{5}\right)}{1-5}+4\right)\)

\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}-1+4\right)\)

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

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

\(=-4\)

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

\(=\sqrt[]{2+2\sqrt[]{2}.1+1}+\left|\sqrt[]{2}-2\right|\)

\(=\sqrt[]{\left(\sqrt[]{2}+1\right)^2}+\left(2-\sqrt[]{2}\right)\) \(\left(\left(\sqrt[]{2}\right)^2=2< 2^2=4\right)\)

\(=\left|\sqrt[]{2}+1\right|+2-\sqrt[]{2}\)

\(=\sqrt[]{2}+1+2-\sqrt[]{2}\)

\(=3\)