Lời giải:
PT tương đương:
\(4\sqrt{x+1}-1=3x+2\sqrt{1-x}+\sqrt{1-x^2}\)
\(\Leftrightarrow 4\sqrt{x+1}-(1+x)=2x+2\sqrt{1-x}+\sqrt{1-x^2}\)
Đặt \(\left\{\begin{matrix} \sqrt{1+x}=a\\ \sqrt{1-x}=b\end{matrix}\right.\Rightarrow 2x=a^2-b^2\)
PT trở thành:
\(4a-a^2=a^2-b^2+2b+ab\)
\(\Leftrightarrow 2a^2+a(b-4)+(2b-b^2)=0\)
\(\Delta=(b-4)^2-8(2b-b^2)=(3b-4)^2\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{4-b+3b-4}{4}=\dfrac{b}{2}\\a=\dfrac{4-b+4-3b}{4}=2-b\end{matrix}\right.\)
TH1: \(a=\frac{b}{2}\Leftrightarrow 2\sqrt{x+1}=\sqrt{1-x}\)
\(\Rightarrow 4(x+1)=1-x\Leftrightarrow x=\frac{-3}{5}\) (thỏa mãn)
TH2: \(a=2-b\Leftrightarrow \sqrt{x+1}=2-\sqrt{1-x}\)
\(\Leftrightarrow \sqrt{x+1}+\sqrt{1-x}=2\)
\(\Rightarrow 2+2\sqrt{1-x^2}=4\Leftrightarrow \sqrt{1-x^2}=1\)
\(\Leftrightarrow x=0\) (thỏa mãn)
Vậy \(x\in\left\{-\frac{3}{5}; 0\right\}\)