a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)

b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

d: \(=\sqrt{5}\left(\sqrt{3}-1\right)-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)

=căn 5-1/2*căn 5

=1/2*căn 5

e: \(=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{1}{\sqrt{6}}\)

f:=2+căn 3+căn 2-2-căn 3=căn 2

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

=18-3

=15

7: Ta có: \(\sqrt{72}+\sqrt{4\dfrac{1}{2}}-\sqrt{32}-\sqrt{162}\)

\(=6\sqrt{2}+\dfrac{3}{2}\sqrt{2}-4\sqrt{2}-9\sqrt{2}\)

\(=-\dfrac{11}{2}\sqrt{2}\)

a) \(2\sqrt{50}-3\sqrt{32}-\sqrt{162}+5\sqrt{98}\)

=\(2.5\sqrt{2}-3.4\sqrt{2}-9\sqrt{2}+5.7\sqrt{2}\)

= \(10\sqrt{2}-12\sqrt{2}-9\sqrt{2}+35\sqrt{2}\)

= \(24\sqrt{2}\)

b) \(\sqrt{8+2\sqrt{7}}+\sqrt{11-4\sqrt{7}}\)

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

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

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

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

c) \(\dfrac{10}{\sqrt{5}}+\dfrac{8}{3+\sqrt{5}}-\dfrac{\sqrt{18}-3\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)

= \(2\sqrt{5}+6-2\sqrt{5}-3\)

= 3

\(A=\dfrac{\sqrt[4]{7\sqrt[3]{54}+15\sqrt[3]{128}}}{\sqrt[3]{\sqrt[4]{32}}+\sqrt[3]{9\sqrt[4]{162}}}\)

\(\Leftrightarrow A=\dfrac{\sqrt[4]{7\sqrt[3]{3^3.2}+15\sqrt[3]{4^3.2}}}{\sqrt[3]{\sqrt[4]{2^4.2}}+\sqrt[3]{9\sqrt[4]{3^4.2}}}\)

\(\Leftrightarrow A=\dfrac{\sqrt[4]{7.3\sqrt[3]{2}+15.4\sqrt[3]{2}}}{\sqrt[3]{2\sqrt[4]{2}}+\sqrt[3]{9.3\sqrt[4]{2}}}\)

\(\Leftrightarrow A=\dfrac{\sqrt[4]{21\sqrt[3]{2}+60\sqrt[3]{2}}}{\sqrt[3]{2\sqrt[4]{2}}+\sqrt[3]{3^3\sqrt[4]{2}}}\)

\(\Leftrightarrow A=\dfrac{\sqrt[4]{81\sqrt[3]{2}}}{\sqrt[3]{\sqrt[4]{2}}\left(\sqrt[3]{2}+3\right)}=\dfrac{3\sqrt[4]{\sqrt[3]{2}}}{\sqrt[3]{\sqrt[4]{2}}\left(\sqrt[3]{2}+3\right)}\)

\(\Leftrightarrow A=\dfrac{3}{\sqrt[3]{2}+3}\)

\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|\sqrt{24}-3\right|=3-\sqrt{6}+\sqrt{24}-3=2\sqrt{6}-\sqrt{6}=\sqrt{6}\)

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=-\dfrac{\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\).

\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=\dfrac{\left(\sqrt{2-\sqrt{3}}\right)^2+\left(\sqrt{2+\sqrt{3}}\right)^2}{\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}=\dfrac{4}{1}=4\)

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

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

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

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

\(a,Sửa:\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\ =\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\\ =2\sqrt{5}-2-2\sqrt{5}=-2\\ b,=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\\ =\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=\dfrac{2\sqrt{6}-\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)