Question

Rút gọn

A=\((\dfrac{1+\sqrt{1-a}}{1-a+\sqrt{1-a}}+\dfrac{1+\sqrt{1+a}}{1+a+\sqrt{1+a}})\cdot\dfrac{a^2-1}{2}+1\)

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Answer 1

\(A=\left(\dfrac{1+\sqrt{1-a}}{1-a+\sqrt{1-a}}+\dfrac{1+\sqrt{1+a}}{1+a+\sqrt{1+a}}\right)\cdot\dfrac{a^2-1}{2}+1\)

\(=\left(\dfrac{1+\sqrt{1-a}}{\sqrt{1-a}\cdot\left(\sqrt{1-a}+1\right)}+\dfrac{1+\sqrt{1+a}}{\sqrt{1-a}\cdot\left(\sqrt{1+a}+1\right)}\right)\cdot\dfrac{a^2-1}{2}+1\)

\(=\left(\dfrac{1}{\sqrt{1-a}}+\dfrac{1}{\sqrt{1+a}}\right)\cdot\dfrac{a^2-1}{2}+1\)

\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{\left(1-a\right)\cdot\left(1+a\right)}}\cdot\dfrac{a^2-1}{2}+1\)

\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1-a^2}}\cdot\dfrac{a^2-1}{2}+1\)