Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\frac{1}{3x} ;\ \ b.\frac{( a-b)( b-a)}{3b} ;\ c.\frac{-1}{6}\\ d.\frac{a}{5} ;\ \ e.\frac{x-1}{x+1}\\ f.\ x< 2\ thì\ A=\frac{x^{2} +x-3}{x-3}\\ 2< x< 3\ thì\ A=\frac{x^{2} -x+1}{x-3} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ \frac{\sqrt{3x}}{\sqrt{27x^{3}}} =\sqrt{\frac{3x}{27x^{3}}} =\sqrt{\frac{1}{( 3x)^{2}}} =|\frac{1}{3x} |=\frac{1}{3x}( \ vì\ x >0)\\ b.\ ( a-b) .\sqrt{\frac{a^{2} -2ab+b^{2}}{9b^{2}}} =( a-b) .|\frac{a-b}{3b} |\\ =\frac{( a-b)( b-a)}{3b}( vì\ b >a >0)\\ c.\ \frac{1}{3xy} .\sqrt{\frac{4x^{2} y^{4}}{16}} =\frac{1}{3xy} .\sqrt{\left(\frac{xy^{2}}{2}\right)^{2}}\\ =\frac{1}{3xy} .|\frac{xy^{2}}{2} |=\frac{-xy}{3xy.2} =\frac{-1}{6}( vì\ x< 0,\ y\neq 0)\\ d.\ \sqrt{\frac{2}{a}} :\sqrt{\frac{50}{a^{3}}} =\sqrt{\frac{2}{a} .\frac{a^{3}}{50}} =\sqrt{\left(\frac{a}{5}\right)^{2}} =|\frac{a}{5} |=\frac{a}{5}( vì\ a >0)\\ e.\ \frac{\sqrt{x^{2} -2x+1}}{\sqrt{x^{2} +2x+1}} =\sqrt{\left(\frac{x-1}{x+1}\right)^{2}} =|\frac{x-1}{x+1} |=\frac{x-1}{x+1}( vì\ x >1)\\ f.\ A=\sqrt{\frac{x^{2} -4x+4}{9-6x+x^{2}}} +\frac{x^{2} -1}{x-3}\\ =\sqrt{\left(\frac{x-2}{3-x}\right)^{2}} +\frac{x^{2} -1}{x-3} =|\frac{x-2}{3-x} |+\frac{x^{2} -1}{x-3}\\ TH1:\ x< 2\ thì\ x-2< 0\ và\ 3-x >1 >0\\ \Leftrightarrow A=\frac{2-x}{3-x} +\frac{x^{2} -1}{x-3} =\frac{x-2+x^{2} -1}{x-3} =\frac{x^{2} +x-3}{x-3}\\ TH2:2< x< 3\ thì\ x-2 >0\ và\ 3-x >0\\ \Leftrightarrow A=\frac{x-2}{3-x} +\frac{x^{2} -1}{x-3} =\frac{2-x+x^{2} -1}{x-3} =\frac{x^{2} -x+1}{x-3} \end{array}$
