$\sqrt{11},\sqrt[{3}]{−16},− \sqrt{27},−\frac{π}{2},3+ \sqrt{29},\frac{\sqrt{2}}{2}$
$−3,\sqrt[{3}]{−16},− \sqrt{27},−\frac{π}{2}$
$解:(1)∵ \sqrt{16}＜ \sqrt{17}＜ \sqrt{25},$ $∴4＜ \sqrt{17}＜5.$ $∴1＜ \sqrt{17}-3＜2.$ $∴a=1,b= \sqrt{17}−4$ $(2)(−a)³+(b+4)²=(−1)²+( \sqrt{17}−4+4)²=−1+17=16.$ $∴ (-a)³+(b+4)²的平方根是±4$
$解:A:−π,E:−\sqrt{5},B:−1.5,D:0.4,F:\sqrt{3},C: \sqrt{10}$ $\sqrt{10}＞\sqrt{3}＞0.4＞−1.5＞−\sqrt{5}＞−π$
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