a) Áp dụng công thức nhị thức Newton, ta có

          \(\begin{array}{l}{\left( {2 + \sqrt 2 } \right)^4} = {2^4} + {4.2^3}.\left( {\sqrt 2 } \right) + {6.2^2}.{\left( {\sqrt 2 } \right)^2} + 4.2.{\left( {\sqrt 2 } \right)^3} + {\left( {\sqrt 2 } \right)^4}\\ = \left[ {{2^4} + {{6.2}^2}.{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^4}} \right] + \left[ {{{4.2}^3}.\left( {\sqrt 2 } \right) + 4.2.{{\left( {\sqrt 2 } \right)}^3}} \right]\\ = 68 + 48\sqrt 2 \end{array}\)

b) Áp dụng công thức nhị thức Newton, ta có

          \({\left( {2 + \sqrt 2 } \right)^4} = {2^4} + {4.2^3}.\left( {\sqrt 2 } \right) + {6.2^2}.{\left( {\sqrt 2 } \right)^2} + 4.2.{\left( {\sqrt 2 } \right)^3} + {\left( {\sqrt 2 } \right)^4}\)

          \({\left( {2 - \sqrt 2 } \right)^4} = \left( {2 +(- \sqrt 2 )} \right)^4= {2^4} + {4.2^3}.\left( { - \sqrt 2 } \right) + {6.2^2}.{\left( { - \sqrt 2 } \right)^2} + 4.2.{\left( { - \sqrt 2 } \right)^3} + {\left( { - \sqrt 2 } \right)^4}\)

Từ đó,

          \(\begin{array}{l}{\left( {2 + \sqrt 2 } \right)^4} + {\left( {2 - \sqrt 2 } \right)^4} = 2\left[ {{2^4} + {{6.2}^2}.{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^4}} \right]\\ = 2\left( {16 + 48 + 4} \right) = 136\end{array}\)

c) Áp dụng công thức nhị thức Newton, ta có

          \(\begin{array}{l}{\left( {1 - \sqrt 3 } \right)^5} = \left( {1 +(- \sqrt 3 )} \right)^5=  1 + 5.\left( { - \sqrt 3 } \right) + 10.{\left( { - \sqrt 3 } \right)^2} + 10.{\left( { - \sqrt 3 } \right)^3} + 5.{\left( { - \sqrt 3 } \right)^4} + 1.{\left( { - \sqrt 3 } \right)^5}\\ = \left[ {1 + 10.{{\left( { - \sqrt 3 } \right)}^2} + 5.{{\left( { - \sqrt 3 } \right)}^4}} \right] + \left[ {5.\left( { - \sqrt 3 } \right) + 10.{{\left( { - \sqrt 3 } \right)}^3} + 1.{{\left( { - \sqrt 3 } \right)}^5}} \right]\\ = 76 - 44\sqrt 3 \end{array}\)

\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)

\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)

\(=\sqrt{29}\)

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

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

\(B=3\cdot1,5-4\cdot\left|3-\sqrt{2}\right|\) \(=4,5-4\left(3-\sqrt{2}\right)\)

\(=4,5-12+4\sqrt{2}\) \(=4\sqrt{2}-7,5\) 

Ta có: \(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)

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

\(=6-\sqrt{7}\)

g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)

=-(căn 5+2)(căn 5-2)

=-(5-4)=-1

h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)

=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6

=4

\(=2\left|3-\sqrt{2}\right|+\sqrt{18}-5.1=6-2\sqrt{2}+3\sqrt{2}-5\)

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

\(D=a^{\dfrac{7}{2}}.a^{\dfrac{1}{3}}.a^{\dfrac{7}{4}}=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}=\sqrt[12]{a^{67}}\)

\(D=a^{\sqrt{2}-1}.a^{2\sqrt{2}}.a^{3-3\sqrt{2}}=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{3}}=a^2\)

\(D=\left(\sqrt{a}\right)^7\cdot\left(\sqrt[3]{a}\right)\left(\sqrt[4]{a}\right)^7\)

\(=a^{\dfrac{1}{2}\cdot7}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}\cdot7}\)

\(=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}\)

b: \(D=a^{\sqrt{2}-1}\cdot\left(a^2\right)^{\sqrt{2}}\cdot\left(a^3\right)^{1-\sqrt{2}}\)

\(=a^{\sqrt{2}-1}\cdot a^{2\sqrt{2}}\cdot a^{3-3\sqrt{2}}\)

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

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

\(=3x+6\sqrt{x}-\left(x-1\right)\)

\(=3x+6\sqrt{x}-x+1\)

\(=2x+6\sqrt{x}+1\)

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

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

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

\(=-x+8\sqrt{x}+1\)

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

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

\(=3x-3\sqrt{x}-2+x-1\)

\(=4x-3\sqrt{x}-3\)

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

\(=x-9-\left(2x-3\sqrt{x}-2\right)\)

\(=x-9-2x+3\sqrt{x}+2\)

\(=-x+3\sqrt{x}-7\)

\(E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)

\(=\sqrt{x^2}-2^2-2\left(2x+4\sqrt{x}-\sqrt{x}-2\right)\)

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

\(=x-4-4x-6\sqrt{x}+4\)

\(=-3-6\sqrt{x}\)

\(1,=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|\\ =1-\sqrt{2}+3+\sqrt{2}\\ =4\\ 2,=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}-2+\sqrt{3}-1\\ =2\sqrt{3}-3\\ 3,=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{5}-3+\sqrt{5}-2\\ =2\sqrt{5}-5\\ 4,=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =3+\sqrt{3}\\ 5,=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\\ =2-\sqrt{3}-\left(2+\sqrt{3}\right)\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)

\(\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)=9-4\left(1+\sqrt{2}\right)=5-4\sqrt{2}\)

=> \(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)=25-16.2=-7\)

=> \(x=-7\sqrt[3]{-7}=7\sqrt[3]{7}\)