Đk: $-1\leq x\leq 1$Pt $\Leftrightarrow 4(\sqrt{x+1}-1)+1-\sqrt{1-x^2}+2(1-\sqrt{1-x})-3x=0$
$\Leftrightarrow 4\frac{x}{\sqrt{x+1}+1}+\frac{x^2}{1+\sqrt{1-x^2}}+2\frac{x}{1+\sqrt{1-x}}-3x=0$
$\Leftrightarrow x(\frac{4}{\sqrt{x+1}+1}+\frac{x}{1+\sqrt{1-x^2}}+\frac{2}{1+\sqrt{1-x}}-3)=0$
$\Leftrightarrow x=0\vee \frac{4}{\sqrt{x+1}+1}-1+\frac{x}{1+\sqrt{1-x^2}}-1+\frac{2}{1+\sqrt{1-x}}-1=0(!)$
$(!)\Leftrightarrow -\frac{3(\sqrt{x+1}-1)}{\sqrt{x+1}+1}+\frac{x-1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}+\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}=0$
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