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a) Ta có: \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
\(=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}-\frac{\sqrt{5}}{2}\)
\(=\frac{2\sqrt{5}-\sqrt{5}}{2}=\frac{\sqrt{5}}{2}\)
b) Ta có: \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{3}-1}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}\)
\(=\frac{\left(2\sqrt{8}-\sqrt{12}\right)\left(\sqrt{3}+1\right)}{3-1}-\frac{\left(\sqrt{5}+\sqrt{27}\right)\left(\sqrt{30}+\sqrt{2}\right)}{30-2}\)
\(=\frac{4\sqrt{6}+4\sqrt{2}-6-2\sqrt{3}}{2}-\frac{5\sqrt{6}+\sqrt{10}+9\sqrt{10}+3\sqrt{6}}{28}\)
\(=\frac{4\sqrt{2}\left(\sqrt{3}+1\right)-2\sqrt{3}\left(\sqrt{3}+1\right)}{2}-\frac{10\sqrt{10}+8\sqrt{6}}{28}\)
\(=\frac{2\cdot\left(\sqrt{3}+1\right)\left(2\sqrt{2}-\sqrt{3}\right)}{2}-\frac{\sqrt{1000}+\sqrt{384}}{28}\)
\(=2\sqrt{6}-3+2\sqrt{2}-\sqrt{3}-\frac{\sqrt{2}\cdot\left(5\sqrt{5}+4\sqrt{3}\right)}{14}\)
\(=\frac{28\sqrt{6}-42+28\sqrt{2}-14\sqrt{3}-10\sqrt{10}-8\sqrt{6}}{14}\)
\(=\frac{20\sqrt{6}-42+28\sqrt{2}-14\sqrt{3}-10\sqrt{10}}{14}\)
c) Ta có: \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}\)
\(=\frac{2\sqrt{3}+2-2\sqrt{3}+2}{2}\)
\(=\frac{4}{2}=2\)
\(A=\sqrt{3}+\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\sqrt{3}+\sqrt{3}\left(\sqrt{3}-1\right)=3\)
\(B=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\sqrt{3}-2-\sqrt{3}=-2\)
\(C=\frac{\sqrt{5}\left(\sqrt{5}+2\right)}{\sqrt{5}}+\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}-\sqrt{5}-\sqrt{3}\)
\(C=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
\(D=\frac{2}{\left|2-\sqrt{5}\right|}-\frac{2}{\left|2+\sqrt{5}\right|}=\frac{2}{\sqrt{5}-2}-\frac{2}{\sqrt{5}+2}=\frac{2\left(\sqrt{5}+2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(D=2\sqrt{5}+4-2\sqrt{5}+4=8\)
\(E=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}-\sqrt{2}=0\)
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
cảm ơn