đkxđ: -1 < x < 1
Đặt: 1 + x = a (a>0) ; 1 - x = b (b>0)
\(M=\dfrac{1+\sqrt{b}}{b+\sqrt{b}}+\dfrac{1-\sqrt{a}}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}}=\dfrac{1+\sqrt{b}}{\sqrt{b}\left(1+\sqrt{b}\right)}-\dfrac{1-\sqrt{a}}{\sqrt{a}\left(1-\sqrt{a}\right)}+\dfrac{1}{\sqrt{a}}=\dfrac{1}{\sqrt{b}}-\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{a}}=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}}+\dfrac{1}{\sqrt{a}}=\dfrac{a+\sqrt{ab}-\sqrt{ab}}{\sqrt{a}\cdot\sqrt{ab}}=\dfrac{a}{a\sqrt{b}}=\dfrac{1}{\sqrt{b}}=\dfrac{1}{\sqrt{1-x}}\)