Đáp án:

$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a.\ |8x|\ ;\ b.\ |3ab|\\ c.\ -a^{2} ;\ d.\ x^{2} -x^{3}\\ Bài\ 2:\\ a.\ 9\ ;\ b.\ 9\\ c.\ 5+2\sqrt{6} ;\ d.\ -7-3\sqrt{3}\\ Bài\ 3:\\ a.\ 3;\ b.\ 0;\ c.\ \sqrt{\frac{a-1}{b-1}} ;\ d.\ x-y \end{array}$

Giải thích các bước giải:

$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a.\ \sqrt{\left( -x^{2}\right) .\left( -8^{2}\right)} =\sqrt{( x.8)^{2}} =|8x|\\ TH1:\ x\geqslant 0\ thì\ |8x|=8x\\ TH2:\ x< 0\ thì\ |8x|=-8x\\ b.\ \sqrt{9a^{2} b^{2}} =\sqrt{( 3ab)^{2}} =|3ab|\\ TH1:ab\geqslant 0\ thì\ |3ab|=3ab\\ TH2:\ ab< 0\ thì\ |3ab|=-3ab\\ c.\ C=\ \frac{1}{a-b}\sqrt{a^{4}( a-b)^{2}} =\frac{1}{a-b} .|a^{2}( a-b) |\\ Với\ a< b\ thì\ a-b< 0\Leftrightarrow |a-b|=b-a\\ \Leftrightarrow C=\frac{a^{2} .( b-a)}{a-b} =\frac{-a^{2}( a-b)}{a-b} =-a^{2}\\ d.D=\ \sqrt{x^{4}( x-1)^{2}} =|x^{2}( x-1) |\\ Với\ x< 1\ thì\ x-1< 0\ \Leftrightarrow |x-1|=1-x\\ D=x^{2}( 1-x) =x^{2} -x^{3}\\ Bài\ 2:\\ a.\ \sqrt{3} .\sqrt{27} =\sqrt{3^{4}} =3^{2}\\ b.\ \sqrt{\sqrt{10} +1} .\sqrt{\sqrt{10} -1} =\sqrt{\left(\sqrt{10} +1\right)\left(\sqrt{10} -1\right)}\\ =\sqrt{10-1} =\sqrt{9} =3^{2}\\ c.\ \left(\sqrt{2} +\sqrt{3}\right)^{2} =2+2.\sqrt{2.3} +3=5+2\sqrt{6}\\ d.\ \left(\sqrt{3} +2\right)\left(\sqrt{3} -5\right) =3-5\sqrt{3} +2\sqrt{3} -10\\ =-7-3\sqrt{3}\\ Bài\ 3:\\ a.\ \frac{\sqrt{27}}{\sqrt{3}} =\sqrt{\frac{27}{3}} =\sqrt{9} =3\\ b.\ \left( 2\sqrt{18} -3\sqrt{32} +6\sqrt{2}\right) :\sqrt{2}\\ =\frac{\sqrt{2} .\left( 2.\sqrt{9} -3.\sqrt{16} +6\right)}{\sqrt{2}} =2.3-3.4+6=0\\ c.\ \sqrt{\frac{\sqrt{a} -1}{\sqrt{b} +1}} :\sqrt{\frac{\sqrt{b} -1}{\sqrt{a} +1}}( a\geqslant 1;\ b >1)\\ =\sqrt{\frac{\sqrt{a} -1}{\sqrt{b} +1}} .\sqrt{\frac{\sqrt{a} +1}{\sqrt{b} -1}} =\sqrt{\frac{\left(\sqrt{a} -1\right)\left(\sqrt{a} +1\right)}{\left(\sqrt{b} +1\right)\left(\sqrt{b} -1\right)}}\\ =\sqrt{\frac{a-1}{b-1}}\\ d.\ \frac{x\sqrt{y} +y\sqrt{x}}{\sqrt{xy}} :\frac{1}{\sqrt{x} -\sqrt{y}} ;\ ĐKXĐ:x >0;\ y >0;\ x\neq y\\ =\frac{\sqrt{xy} .\left(\sqrt{x} +\sqrt{y}\right)}{\sqrt{xy}} .\left(\sqrt{x} -\sqrt{y}\right) =\left(\sqrt{x} +\sqrt{y}\right) .\left(\sqrt{x} -\sqrt{y}\right)\\ =x-y \end{array}$