Giải câu 5 trang 34 sách phát triển năng lực toán 9 tập 1.
P = $(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^{2}}-1+a})(\sqrt{\frac{1}{a^{2}}-1}-\frac{1}{a})$
= $(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{(\sqrt{1-a})^{2}}{\sqrt{(1-a)(1+a)}-(\sqrt{1-a})^{2}})(\sqrt{\frac{1-a^{2}}{a^{2}}}-\frac{1}{a})$
= $(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{(\sqrt{1-a})^{2}}{\sqrt{1-a}(\sqrt{1+a}-\sqrt{1-a})})(\sqrt{\frac{(1-a)(1+a)}{a^{2}}}-\frac{1}{a})$
= $(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}})(\frac{\sqrt{1-a}.\sqrt{1+a}}{a}-\frac{1}{a})$
= $\frac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\frac{2\sqrt{1-a}\sqrt{1+a}-(1-a)-(1+a)}{2a}$
= $\frac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\frac{-(\sqrt{1+a}-\sqrt{1-a})^{2}}{2a}$
=$-\frac{(\sqrt{1+a}+\sqrt{1-a})(\sqrt{1+a}-\sqrt{1-a})}{2a}=-\frac{1+a-1+a}{2a}=-\frac{2a}{2a}=-1$