\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

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

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

\(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}=\sqrt{\frac{3}{7}}\)

\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)

\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)

\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)

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

\(C=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(C=\sqrt{6+2.\left(\sqrt{3}-1\right)}\)

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

\(C=\sqrt{4+2\sqrt{3}}\)

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

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

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

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

         \(=\sqrt{2}+1+\sqrt{2}-1\)

         \(=2\sqrt{2}\approx2,82843\)

2) Ta có: \(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)

        \(\Leftrightarrow B=\frac{\sqrt{5}.\sqrt{3}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\)

        \(\Leftrightarrow B=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}\)

        \(\Leftrightarrow B=\frac{\sqrt{3}}{\sqrt{7}}\approx0,65465\)

3) Ta có: \(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)

        \(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}\)

        \(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{8}.\sqrt{3-\sqrt{3}-1}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{2.8-2.2.\sqrt{3}.2}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{4.3}.2+1}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{12}.2+4}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{\left(\sqrt{12}-2\right)^2}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{12}-2}\)

        \(\Leftrightarrow C=\sqrt{3+2\sqrt{3}+1}\)

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

        \(\Leftrightarrow C=\sqrt{3}+1\approx2,73205\)

a) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

b)\(\frac{x-4}{2\left(\sqrt{x}+2\right)}\) (ĐK:x\(\ge0\))

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

\(=\frac{\sqrt{x}-2}{2}\)

c)\(\frac{x-5\sqrt{x}+6}{3\sqrt{x}-6}\) (ĐK:x\(\ge0;x\ne4\))

\(=\frac{x-3\sqrt{x}-2\sqrt{x}+6}{3\left(\sqrt{x}-2\right)}\)

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

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

\(=\frac{\sqrt{x}-3}{3}\)

a)\(\frac{3\sqrt{6}-\sqrt{2}}{1-3\sqrt{3}}=\frac{3\sqrt{3}.\sqrt{2}-\sqrt{2}}{1-3\sqrt{3}}=\frac{\sqrt{2}.\left(3\sqrt{3}-1\right)}{-\left(3\sqrt{3}-1\right)}=-\sqrt{2}\)

b)\(\frac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}=\frac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2.\left(\sqrt{2}-\sqrt{3}\right)}=\frac{\sqrt{5}}{2}\)

c)\(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}\)

d)\(\frac{5\sqrt{6}-6\sqrt{5}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{5^2.6}-\sqrt{6^2.5}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{30}.\sqrt{5}-\sqrt{30}.\sqrt{6}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{30}.\left(\sqrt{5}-\sqrt{6}\right)}{\sqrt{5}-\sqrt{6}}=\sqrt{30}\)

e)\(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{6}}=\frac{\sqrt{2^2.3}-\sqrt{3^2.2}}{\sqrt{6}}=\frac{\sqrt{6}.\sqrt{2}-\sqrt{6}.\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{6}.\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}}=\sqrt{2}-\sqrt{3}\)

f)\(\frac{6\sqrt{2}-4}{\sqrt{2}}=\frac{6\sqrt{2}-\sqrt{16}}{\sqrt{2}}=\frac{6\sqrt{2}-\sqrt{2}.2\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}.\left(6-2\sqrt{2}\right)}{\sqrt{2}}=6-2\sqrt{2}\)

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

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