Rút gọn các biểu thức sau:
a) \({2^{\sqrt 3 + 1}}:{2^{\sqrt 3 - 1}}\);
b) \({\left( {{3^{\sqrt 2 }}} \right)^{\sqrt 8 }}\);
c) \({\left[ {{{\left( {\sqrt 7 } \right)}^{\sqrt 2 }}} \right]^{\sqrt 8 }}\);
d) \({a^{2\sqrt 5 + 1}}:{a^{2\sqrt 5 - 2}}\);
e) \({3^{3 + \sqrt 2 }}{.3^{ - 1 + \sqrt 2 }}{.9^{1 - \sqrt 2 }}\);
g) \({\left( {{a^{ - \sqrt 3 }}.{b^{\frac{1}{{\sqrt 3 }}}}} \right)^{\frac{1}{{\sqrt 3 }}}}\).
Sử dụng các công thức về phép tính lũy thừa để tính:
a, d) \(\frac{{{a^\alpha }}}{{{a^\beta }}} = {a^{\alpha - \beta }}\)
b, c) \({\left( {{a^\alpha }} \right)^\beta } = {a^{\alpha \beta }}\)
e) \({a^\alpha }.{a^\beta } = {a^{\alpha + \beta }}\), \({\left( {{a^\alpha }} \right)^\beta } = {a^{\alpha \beta }}\)
g) \({\left( {a.b} \right)^\alpha } = {a^\alpha }.{b^\alpha }\)
a) \({2^{\sqrt 3 + 1}}:{2^{\sqrt 3 - 1}} = {2^{\sqrt 3 + 1 - \sqrt 3 + 1}} = {2^2} = 4\);
b) \({\left( {{3^{\sqrt 2 }}} \right)^{\sqrt 8 }} = {3^{\sqrt 2 .\sqrt 8 }} = {3^4} = 81\);
c) \({\left[ {{{\left( {\sqrt 7 } \right)}^{\sqrt 2 }}} \right]^{\sqrt 8 }} = {\left( {\sqrt 7 } \right)^{\sqrt 2 .\sqrt 8 }} = {\left( {\sqrt 7 } \right)^4} = 49\);
d) \({a^{2\sqrt 5 + 1}}:{a^{2\sqrt 5 - 2}} = {a^{2\sqrt 5 + 1 - 2\sqrt 5 + 2}} = {a^3}\);
e) \({3^{3 + \sqrt 2 }}{.3^{ - 1 + \sqrt 2 }}{.9^{1 - \sqrt 2 }} = {3^{3 + \sqrt 2 + - 1 + \sqrt 2 }}{.3^{2\left( {1 - \sqrt 2 } \right)}} = {3^{2 + 2\sqrt 2 }}{.3^{2 - 2\sqrt 2 }} = {3^{2 + 2\sqrt 2 + 2 - 2\sqrt 2 }} = {3^4} = 81\);
g) \({\left( {{a^{ - \sqrt 3 }}.{b^{\frac{1}{{\sqrt 3 }}}}} \right)^{\frac{1}{{\sqrt 3 }}}} = {a^{ - \sqrt 3 .\frac{1}{{\sqrt 3 }}}}.{b^{\frac{1}{{\sqrt 3 }}.\frac{1}{{\sqrt 3 }}}} = \frac{1}{a}.{b^{\frac{1}{3}}} = \frac{{\sqrt[3]{b}}}{a}\).