Bài 2:
\(A=\left(3-\sqrt{3}\right)\left(-2\sqrt{3}\right)+\left(3\sqrt{3}+1\right)^2\)
\(=-6\sqrt{3}+6+28+6\sqrt{3}\)
=28+6
=34
\(B=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt[]{2}+1}-\left(2+\sqrt{3}\right)\)
\(=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
\(=2+\sqrt{3}+\sqrt{2}-\left(2+\sqrt{3}\right)=\sqrt{2}\)
\(C=\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=\sqrt{2}-1-2-\sqrt{2}=-3\)
Bài 3:
\(E=\left(\dfrac{3}{2}\sqrt{6}+2\cdot\sqrt{\dfrac{2}{3}}-4\cdot\sqrt{\dfrac{3}{2}}\right)\left(3\sqrt{\dfrac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\cdot\left(-\sqrt{2}\right)\)
\(=\left(\dfrac{3}{2}\sqrt{6}+2\cdot\dfrac{\sqrt{6}}{3}-4\cdot\dfrac{\sqrt{6}}{2}\right)\left(3\cdot\dfrac{\sqrt{6}}{3}-2\sqrt{3}-\sqrt{6}\right)\cdot\left(-\sqrt{2}\right)\)
\(=\dfrac{1}{6}\sqrt{6}\cdot\left(-2\sqrt{3}\right)\cdot\left(-\sqrt{2}\right)=2\sqrt{6}\cdot\dfrac{1}{6}\sqrt{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
\(F=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}\)
\(H=\sqrt{8+\sqrt{60}}+\sqrt{45}-\sqrt{12}\)
\(=\sqrt{8+2\sqrt{15}}+3\sqrt{5}-2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}\)
\(=\sqrt{5}+\sqrt{3}+3\sqrt{5}-2\sqrt{3}=4\sqrt{5}-\sqrt{3}\)
\(I=\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\sqrt{5}-2-\left(\sqrt{5}+2\right)=\sqrt{5}-2-\sqrt{5}-2=-2-2=-4\)
\(M=\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
=3-2=1
\(N=\dfrac{3+\sqrt{5}}{3-\sqrt{5}}+\dfrac{3-\sqrt{5}}{3+\sqrt{5}}\)
\(=\dfrac{\left(3+\sqrt{5}\right)^2+\left(3-\sqrt{5}\right)^2}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=\dfrac{14+6\sqrt{5}+14-6\sqrt{5}}{9-5}=\dfrac{28}{4}=7\)
