第321题

参考答案：\(\sqrt[6]{{8\sqrt 2 }} = \sqrt[6]{{{2^3} \cdot {2^{\frac{1}{2}}}}} = {({2^{\frac{7}{2}}})^{\frac{1}{6}}} = {2^{\frac{7}{{12}}}};\)

第322题

参考答案：\(\sqrt {a\sqrt a } = \sqrt {a \cdot {a^{\frac{1}{2}}}} = \sqrt {{a^{\frac{3}{2}}}} = {({a^{\frac{3}{2}}})^{\frac{1}{2}}} = {a^{\frac{3}{4}}};\)

第323题

参考答案：\({b^3} \cdot \sqrt[3]{{{b^2}}} = {b^3} \cdot {b^{\frac{2}{3}}} = {b^{\frac{{11}}{3}}};\)

第324题

参考答案：\(\frac{1}{{\sqrt[3]{{x{{(\sqrt[5]{{{x^2}}})}^2}}}}} = \frac{1}{{\sqrt[3]{{x \cdot {{({x^{\frac{2}{5}}})}^2}}}}} = \frac{1}{{\sqrt[3]{{x \cdot {x^{\frac{4}{5}}}}}}} = \frac{1}{{\sqrt[3]{{{x^{\frac{9}{5}}}}}}} = \frac{1}{{{{({x^{\frac{9}{5}}})}^{\frac{1}{3}}}}} = \frac{1}{{{x^{\frac{3}{5}}}}} = {x^{ - \frac{3}{5}}}.\)

第325题

参考答案：

第326题

参考答案：法一：\(\sqrt {\frac {3-2\sqrt {2}} {3+2\sqrt {2}}}=\sqrt {\frac {\left ( {\sqrt {2}-1} \right )^{2}} {\left ( {\sqrt {2}+1} \right )^{2}}}=\frac {\sqrt {2}-1} {\sqrt {2}+1}=\frac {\left ( {\sqrt {2}-1} \right )^{2}} {\left ( {\sqrt {2}+1} \right )\left ( {\sqrt {2}-1} \right )}=3-2\sqrt {2}\)

法二：\(\sqrt {\frac {3-2\sqrt {2}} {3+2\sqrt {2}}}=\sqrt {\frac {\left ( {3-2\sqrt {2}} \right )^{2}} {\left ( {3+2\sqrt {2}} \right )\left ( {3-2\sqrt {2}} \right )}}=3-2\sqrt {2}\)

第327题

参考答案：∵\(\sqrt[{3}] {\left ( {-6} \right )^{3}}=-6，\sqrt[{4}] {\left ( {\sqrt {5}-4} \right )^{4}}=\left | {\sqrt {5}-4} \right |=4-\sqrt {5}，\sqrt[{3}] {\left ( {\sqrt {5}-4} \right )^{3}}=\sqrt {5}-4\) ，

∴原式\(-6+4-\sqrt {5}+\sqrt {5}-4=6\) .

第328题

参考答案：原式 \( = \sqrt {{{(\sqrt 3 )}^2} + 2\sqrt 6 + {{(\sqrt 2 )}^2}} + \sqrt {{{(\sqrt 3 )}^2} - 2\sqrt 6 + {{(\sqrt 2 )}^2}} \)\( = \sqrt {{{(\sqrt 3 + \sqrt 2 )}^2}} + \sqrt {{{(\sqrt 3 - \sqrt 2 )}^2}} = |\sqrt 3 + \sqrt 2 | + |\sqrt 3 - \sqrt 2 |\)\( = \sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 = 2\sqrt 3 \).

第329题

参考答案：\({3^{\frac{1}{4}}}\)；\(9\sqrt 5 .\)

第330题

参考答案：原式= \({[{(0.3)^3}]^{\tfrac{2}{3}}} + {[{(\frac{3}{5})^3}]^{ - \tfrac{1}{3}}} - {[{(\frac{5}{3})^2}]^{0.5}}\)
= \({(0.3)^2} + {(\frac{3}{5})^{ - 1}} - \frac{5}{3}\)
= \(\frac{9}{{100}} + \frac{5}{3} - \frac{5}{3}\)
= \(\frac{9}{{100}}\)

第331题

参考答案：原式=\(\left ( {-2{a}^{\frac {1} {3}}{b}^{-\frac {3} {4}}} \right )\cdot {a}^{3}{b}^{-2}\div 4{a}^{\frac {10} {3}}{b}^{-\frac {11} {4}}
=\left [ {\left ( {-2} \right )\div 4} \right ]{\cdot a}^{\frac {1} {3}+3-\frac {10} {3}}\cdot {b}^{-\frac {3} {4}-2+\frac {11} {4}}
=-\frac {1} {2}\)

第332题

A.\(\frac{1}{{{6^4}}}\)

B.\({2^{2n + 5}}\)

C.\({2^{{n^2} - 2n + 6}}\)

D.\({\left( {\frac{1}{2}} \right)^{2n - 7}}\)

参考答案：D

第333题

参考答案：\(\frac{1}{{2(1 - {2^{ - \tfrac{1}{{32}}}})}}\)

第334题

参考答案：①\(\because {({a^{\tfrac{1}{2}}} + {a^{ - \tfrac{1}{2}}})^2} = 9\)
\(\therefore a + {a^{ - 1}} + 2 = 9\)
\(\therefore a + {a^{ - 1}} = 7\)

第335题

参考答案：②\(\because {(a + {a^{ - 1}})^2} = {a^2} + {a^{ - 2}} + 2 = 49\)\(\therefore {a^2} + {a^{ - 2}} = 47\)

第336题

参考答案：③原式= \(\frac {\left ( {{a}^{\frac {1} {2}}-{a}^{-\frac {1} {2}}} \right )\left ( {a+{a}^{\frac {1} {2}}\cdot {a}^{-\frac {1} {2}}+{a}^{-1}} \right )} {{a}^{\frac {1} {2}}-{a}^{-\frac {1} {2}}}\)

= \(a + {a^{ - 1}} + 1\)

= \(7 + 1\)

= \(8\)

第337题

参考答案：8；1

第338题

A.\(\left( {1, + \infty } \right)\)

B.\(\left( {1,8} \right)\)

C.\(\left( {4,8} \right)\)

D.\(\left[ {4,8} \right)\)

参考答案：D

第339题

A.\(\left( { - \infty ,0} \right)\)

B.\(\left( {1,2} \right)\)

C.\(\left( {0, + \infty } \right)\)

D.\(\left( {0,1} \right)\)

参考答案：D

第340题

A.\(( - 2, - 1)\)

B.\(( - 2, + \infty )\)

C.\(\left( { - \infty , - 1} \right]\)

D.\(\left( { - 2, - 1} \right]\)

参考答案：D