Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ Q=\frac{\sqrt{a-b}}{\sqrt{a+b}}\\ b.\ Q=\sqrt{\frac{1}{2}} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Q=\frac{a}{\sqrt{a^{2} -b^{2}}} -\left( 1+\frac{a}{\sqrt{a^{2} -b^{2}}}\right) :\frac{b}{a-\sqrt{a^{2} -b^{2}}}( a >b >0)\\ a.\ A=\frac{a}{\sqrt{a^{2} -b^{2}}} -\left(\frac{\sqrt{a^{2} -b^{2}} +a}{\sqrt{a^{2} -b^{2}}}\right) .\frac{a-\sqrt{a^{2} -b^{2}}}{b}\\ =\frac{ab-\left( a^{2} -\left(\sqrt{a^{2} -b^{2}}\right)^{2}\right)}{b.\sqrt{a^{2} -b^{2}}} =\frac{ab-a^{2} +a^{2} -b^{2}}{b\sqrt{a^{2} -b^{2}}}\\ =\frac{b( a-b)}{b\sqrt{( a-b)( a+b)}} =\frac{\sqrt{a-b}}{\sqrt{a+b}}\\ b.\ Khi\ a=3b\ thì\ Q=\frac{\sqrt{3b-b}}{\sqrt{3b+b}} =\sqrt{\frac{2b}{4b}} =\sqrt{\frac{1}{2}} \end{array}$
