Điều kiện xác định $1-x^2\ge 0\Rightarrow |x|\le 1$
$\begin{array}{l} \sqrt {x + 1} + 2\left( {x + 1} \right) = x - 1 + \sqrt {1 - x} + 3\sqrt {1 - {x^2}} \\ \Leftrightarrow \sqrt {x + 1} + 2\left( {x + 1} \right) - 3\sqrt {1 - {x^2}} + 1 - x - \sqrt {1 - x} = 0\\ \Leftrightarrow 2\left( {x + 1} \right) - \sqrt {1 - {x^2}} + \sqrt {x + 1} - 2\sqrt {1 - {x^2}} + 1 - x - \sqrt {1 - x} = 0\\ \Leftrightarrow \sqrt {x + 1} \left( {2\sqrt {x + 1} - \sqrt {1 - x} + 1} \right) - \sqrt {1 - x} \left( {2\sqrt {x + 1} - \sqrt {1 - x} + 1} \right) = 0\\ \Leftrightarrow \left( {2\sqrt {x + 1} - \sqrt {1 - x} + 1} \right)\left( {\sqrt {x + 1} - \sqrt {1 - x} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 2\sqrt {x + 1} - \sqrt {1 - x} + 1 = 0\\ \sqrt {x + 1} - \sqrt {1 - x} = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt {x + 1} = \sqrt {1 - x} \\ 2\sqrt {x + 1} = \sqrt {1 - x} - 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x + 1 = 1 - x\\ 4\left( {x + 1} \right) = 1 - x + 1 - 2\sqrt {1 - x} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 2x = 0\\ 4x + 4 = 2 - x - 2\sqrt {1 - x} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 5x + 2 + 2\sqrt {1 - x} = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 5x + 2 = - 2\sqrt {1 - x} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ {\left( {5x + 2} \right)^2} = 4\left( {1 - x} \right) \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 25{x^2} + 20x + 4 = 4 - 4x \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 25{x^2} + 24x = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = - \dfrac{{24}}{{25}} \end{array} \right. \end{array}$
Thử lại ta có hai nghiệm $x=0, x=-\dfrac{24}{25}$ thỏa mãn phương trình.
