a: =2-căn 3-2-căn 3

=-2căn 3

b: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}+1-\sqrt{7}+1\right)=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

c: \(A=\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)

=>\(A^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}+2\cdot\sqrt{16-10+2\sqrt{5}}\)

\(\Leftrightarrow A^2=8+2\left(\sqrt{5}+1\right)=10+2\sqrt{5}\)

=>\(A=\sqrt{10+2\sqrt{5}}\)

b) làm thế nào để ra được \(\dfrac{1}{ \sqrt{2}}\)\((\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}})\) vậy ạ?????

Ta có: \(A=\frac{-\left(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\right)}{\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}}\)

\(=\frac{-\left(\sqrt{3-2\cdot\sqrt{3}\cdot2+4}+\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\right)}{\sqrt{3-2\cdot\sqrt{3}\cdot2+4}-\sqrt{3+2\cdot\sqrt{3}\cdot2+4}}\)

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

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

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

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

\(T=\frac{\sqrt{2}.\left(4+\sqrt{7}\right)}{\sqrt{2}.\left(2\sqrt{2}+\sqrt{4+\sqrt{7}}\right)}+\frac{\sqrt{2}.\left(4-\sqrt{7}\right)}{\sqrt{2}.\left(2\sqrt{2}-\sqrt{4-\sqrt{7}}\right)}\)

\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\sqrt{8+2\sqrt{7}}}+\frac{4\sqrt{2}-\sqrt{14}}{4-\sqrt{8-2\sqrt{7}}}\)

\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\sqrt{7+2\sqrt{7}+1}}+\frac{4\sqrt{2}-\sqrt{14}}{4-\sqrt{7-2\sqrt{7}+1}}\)

\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\left(\sqrt{7}+1\right)^2}+\frac{4\sqrt{2}-\sqrt{14}}{4-\left(\sqrt{7}-1\right)^2}\)\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+|\sqrt{7}+1|}+\frac{4\sqrt{2}-\sqrt{14}}{4-|\sqrt{7}-1|}\)

\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\sqrt{7}+1}+\frac{4\sqrt{2}-\sqrt{14}}{4-\sqrt{7}+1}\)

\(T=\frac{4\sqrt{2}+\sqrt{14}}{5+\sqrt{7}}+\frac{4\sqrt{2}-\sqrt{14}}{5-\sqrt{7}}\)

\(T=\frac{\left(4\sqrt{2}+\sqrt{14}\right).\left(5-\sqrt{7}\right)}{\left(5+\sqrt{7}\right).\left(5-\sqrt{7}\right)}+\frac{\left(4\sqrt{2}-\sqrt{14}\right).\left(5+\sqrt{7}\right)}{\left(5+\sqrt{7}\right).\left(5-\sqrt{7}\right)}\)

\(T=\frac{20\sqrt{2}-\sqrt{98}}{9}\)

\(T=\frac{13\sqrt{2}}{9}\)

B = \(\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)

=>  \(\frac{2}{\sqrt{2}}B=\frac{8+2\sqrt{7}}{6+\sqrt{8+2\sqrt{7}}}+\frac{8-2\sqrt{7}}{6-\sqrt{8-2\sqrt{7}}}\)

=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{6+\sqrt{7}+1}+\frac{\left(\sqrt{7}-1\right)^2}{6-\sqrt{7}+1}\)

=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{\sqrt{7}\left(\sqrt{7}+1\right)}+\frac{\left(\sqrt{7}-1\right)^2}{\sqrt{7}\left(\sqrt{7}-1\right)}\)

=> \(\frac{2}{\sqrt{2}}B=\frac{\sqrt{7}+1}{\sqrt{7}}+\frac{\sqrt{7}-1}{\sqrt{7}}=\frac{2\sqrt{7}}{\sqrt{7}}=2\)

=> B = \(\sqrt{2}\)

\(B=\frac{1}{\sqrt{5}+\sqrt{7}}-\frac{1}{\sqrt{5}-\sqrt{7}}=\frac{\sqrt{5}-\sqrt{7}-\sqrt{5}-\sqrt{7}}{5-7}=\frac{-2\sqrt{7}}{-2}=\sqrt{7}\)

\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}=\sqrt{\left(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)^2}\)

\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}+2\sqrt{\frac{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}}+\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)

\(C=\sqrt{\frac{\left(4+\sqrt{7}\right)^2}{16-7}+\frac{\left(4-\sqrt{7}\right)^2}{16-7}+2}\)

\(C=\sqrt{\frac{\left(4+\sqrt{7}+4-\sqrt{7}\right)^2-2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{16-7}+2}\)

\(C=\sqrt{\frac{16^2-2\left(16-7\right)}{9}+2}=\sqrt{\frac{238}{9}+2}=\sqrt{\frac{256}{9}}=\frac{16}{3}\)

Chúc bạn học tốt ~ 

thanks ban 

3: \(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

4: \(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\sqrt{2}\)

5: \(=\dfrac{\sqrt{23-8\sqrt{7}}}{3}+\dfrac{\sqrt{23+8\sqrt{7}}}{3}\)

\(=\dfrac{4-\sqrt{7}+4+\sqrt{7}}{3}=\dfrac{8}{3}\)

\(B=\frac{\sqrt{2}\left(4+\sqrt{7}\right)}{6+\sqrt{8+2\sqrt{7}}}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)}{6-\sqrt{8-2\sqrt{7}}}=\frac{\sqrt{2}\left(4+\sqrt{7}\right)}{6+\sqrt{\left(\sqrt{7}+1\right)^2}}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)}{6-\sqrt{\left(\sqrt{7}-1\right)^2}}\)

\(=\frac{\sqrt{2}\left(4+\sqrt{7}\right)}{7+\sqrt{7}}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)}{7-\sqrt{7}}=\frac{\sqrt{2}\left(4+\sqrt{7}\right)\left(7-\sqrt{7}\right)}{35}+\frac{\sqrt{2}\left(4-\sqrt{7}\right)\left(7+\sqrt{7}\right)}{35}\)

\(=\frac{21\sqrt{2}+3\sqrt{14}}{35}+\frac{21\sqrt{2}-3\sqrt{14}}{35}=\frac{42\sqrt{2}}{35}=\frac{6\sqrt{2}}{5}\)

\(a.\left(4+\sqrt{7}\right)\left(\sqrt{14}-\sqrt{2}\right)\sqrt{4-\sqrt{7}}=\left(4+\sqrt{7}\right)\left(\sqrt{7}-1\right)\sqrt{7-2\sqrt{7}+1}=\left(4+\sqrt{7}\right)\left(\sqrt{7}-1\right)^2=2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)=2\left(16-7\right)=18\) \(b.\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}=\dfrac{4\sqrt{2}+\sqrt{14}}{6+\sqrt{7+2\sqrt{7}+1}}+\dfrac{4\sqrt{2}-\sqrt{14}}{6-\sqrt{7-2\sqrt{7}+1}}=\dfrac{4\sqrt{2}+\sqrt{14}}{7+\sqrt{7}}+\dfrac{4\sqrt{2}-\sqrt{14}}{7-\sqrt{7}}=\dfrac{\left(4\sqrt{2}+\sqrt{14}\right)\left(7-\sqrt{7}\right)+\left(4\sqrt{2}-\sqrt{14}\right)\left(7+\sqrt{7}\right)}{49-7}=\dfrac{28\sqrt{2}-4\sqrt{14}+7\sqrt{14}-7\sqrt{2}+28\sqrt{2}+4\sqrt{14}-7\sqrt{14}-7\sqrt{2}}{42}=\dfrac{42\sqrt{2}}{42}=\sqrt{2}\)