Giải câu 1 đề 13 ôn thi toán 9 lên 10.
a. $A=\frac{a\sqrt{a}-1}{a-\sqrt{a}} -\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left [ \sqrt{a}-\frac{1}{\sqrt{a}} \right ]\left [ \frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1} \right ]$
$=\frac{(\sqrt{a}-1)(a+\sqrt{a}+1)}{\sqrt{a}(\sqrt{a}-1)}-\frac{(\sqrt{a}+1)(a-\sqrt{a}+1)}{\sqrt{a}(\sqrt{a}+1)}+\left [ \sqrt{a}-\frac{1}{\sqrt{a}} \right ]\left [ \frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1} \right ]$
$=\frac{(a+\sqrt{a}+1)-(a-\sqrt{a}+1)}{\sqrt{a}}+\frac{a-1}{\sqrt{a}}.\frac{(\sqrt{a}+1)^{2}+(\sqrt{a}-1)^{2}}{(\sqrt{a}-1)(\sqrt{a}+1)}$ $=2+\frac{2a+2}{\sqrt{a}}$
b. $A> 6\Rightarrow 2+\frac{2a+2}{\sqrt{a}}>6$
$\Leftrightarrow \frac{2aa+2}{\sqrt{a}}-4>0\Leftrightarrow \frac{a-2\sqrt{a}+1}{\sqrt{a}}>0\Leftrightarrow \frac{(\sqrt{a}-1)^{2}}{\sqrt{a}}>0$
$\Rightarrow \left\{\begin{matrix}(\sqrt{a}-1)^{2}>0& & \\ \sqrt{a}>0& & \end{matrix}\right.$ luôn đúng với a > 0, a # 1
Vậy với a > 0, a # 1 thì A > 6