Cho biểu thức P=(căn(1+a)/(căn(1+a)−căn(1−a))+(1−a)/(căn(1−a^2)−1+a))(căn(1/a^2)−1)−1/a) với 0 < a < 1

* Hướng dẫn giải

Với 0 < a < 1 ta có:

$\begin{array}{l}P=\left[\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{{\left(\sqrt{1-a}\right)}^2}{\sqrt{\left(1-a\right)\left(1+a\right)}-{\left(\sqrt{1-a}\right)}^2}\right]\left(\sqrt{\frac{1-a^2}{a^2}}-\frac{1}{a}\right)\\ =\left[\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{{\left(\sqrt{1-a}\right)}^2}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right]\left[\sqrt{\frac{(1-a)(1+a)}{a^2}}-\frac{1}{a}\right]\\ =\left[\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right]\left(\frac{\sqrt{1-a}.\sqrt{1+a}}{a^2}-\frac{1}{a}\right)\\ =\frac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\frac{2\sqrt{1-a}.\sqrt{1+a}-(1-a)-(1+a)}{2a}\\ =\frac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\frac{-{\left(\sqrt{1+a}-\sqrt{1-a}\right)}^2}{2a}\\ =-\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)\left(\sqrt{1+a}-\sqrt{1-a}\right)}{2a}\\ =-\frac{1+a-1+a}{2a}=-\frac{2a}{2a}=-1\end{array}$