Điều kiện: $-1\le x\le 1$
Ta tìm $\alpha,\beta$ sao cho:
$-x+3=\alpha(1+x)+\beta(1-x)\Leftrightarrow \left\{\begin{array}{l}\alpha-\beta=-1 \\\alpha+\beta=3 \end{array}\right. \Leftrightarrow \left\{\begin{array}{l}\alpha=1\\\beta=2\end{array}\right.$
Khi đó phương trình viết lại như sau:
$(1+x)+2(1-x)-2\sqrt{1-x}+\sqrt{1+x}-3\sqrt{1-x^2}=0$
Đặt $u=\sqrt{1+x},v=\sqrt{1-x},(u,v\ge0)$, ta được:
$u^2+2v^2-2v+u-3uv=0$
$\Leftrightarrow (u-2v)(u-v+1)=0$
$\Leftrightarrow \left[\begin{array}{l}u=2v\\u-v+1=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\sqrt{1+x}=2\sqrt{1-x}\\\sqrt{1+x}+1=\sqrt{1-x}\end{array}\right. \Leftrightarrow \left[\begin{array}{l}x=\frac{3}{5}\\x=\frac{-\sqrt{3}}{2}\end{array}\right.$