Question

Cho biểu thức \(P=\left(\frac{1}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{1}{\sqrt{1-a^2}}+1\right)\)

a)Rút gọn P

b)Tính P khi \(a=\frac{\sqrt{3}}{2+\sqrt{3}}\)

Answer 1

\(a.P=\left(\frac{1}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{1}{\sqrt{1-a^2}}+1\right)=\left(\frac{\sqrt{1-a}+\sqrt{1-a}.\sqrt{1-a^2}}{\sqrt{1-a^2}}\right):\left(\frac{1+\sqrt{1-a^2}}{\sqrt{1-a^2}}\right)=\frac{\sqrt{1-a}.\left(1+\sqrt{1-a^2}\right)}{1+\sqrt{1-a^2}}=\sqrt{1-a}\)

Answer 2

\(P=\sqrt{1-\frac{\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\frac{2+\sqrt{3}-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\frac{2}{2+\sqrt{3}}}=\sqrt{\frac{2.2}{2.\left(2+\sqrt{3}\right)}}=\sqrt{\frac{4}{4+2\sqrt{3}}}=\sqrt{\frac{4}{3+2\sqrt{3}+1}}=\sqrt{\frac{4}{\left(\sqrt{3}+1\right)^2}}=\frac{2}{\left|\sqrt{3}+1\right|}=\frac{2\left(\sqrt{3}-1\right)}{3-1}=\sqrt{3}-1\)