a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

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

b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)

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

\(=1-5-2\sqrt{6}\)

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

\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)

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

\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)

\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)

a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)

\(=\sqrt{5}-1\)

b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)

\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)

\(=2\sqrt{2}\)

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

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

a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)

b) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)

c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)

d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)

e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)

f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)

a) Ta có: \(D=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\cdot\left(-\sqrt{2}\right)\)

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

\(=-2+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\)

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

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

\(=-2+\sqrt{5}-1\)(Vì \(\sqrt{5}>1\))

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

b) Ta có: \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}\right)-\sqrt{75}\)

\(=2\sqrt{81}+4\sqrt{144}-5\sqrt{3}\)

\(=18+48-5\sqrt{3}\)

\(=66-5\sqrt{3}\)

c) Ta có: \(E=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)

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

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

\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\left|\sqrt{5}-\sqrt{3}\right|\)

\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))

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

\(=2\sqrt{2}\)

d) Ta có: \(P=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

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

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

\(=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\)(Vì \(\sqrt{\frac{3}{2}}>\sqrt{\frac{1}{2}}>0\))

\(=2\sqrt{\frac{3}{2}}=\sqrt{4\cdot\frac{3}{2}}=\sqrt{6}\)

e) Ta có: \(M=-3\sqrt{50}+2\sqrt{98}-7\sqrt{72}\)

\(=\sqrt{2}\cdot\left(-3\cdot\sqrt{25}+2\cdot\sqrt{49}-7\cdot\sqrt{36}\right)\)

\(=\sqrt{2}\cdot\left(-15+14-42\right)\)

\(=-43\sqrt{2}\)