Giải thích các bước giải:
ĐKXĐ : $-1\le x\le 1$
$\sqrt{x+1}+2(x+1)=(x-1)+\sqrt{1-x}+3\sqrt{1-x^2}$
$\rightarrow 2(x+1)-3\sqrt{1-x}.\sqrt{1+x}+(1-x)+(\sqrt{x+1}+\sqrt{1-x})=0$
$\rightarrow (\sqrt{x+1}-\sqrt{1-x})(2\sqrt{x+1}-\sqrt{1-x})+(\sqrt{x+1}-\sqrt{1-x})=0$
$\rightarrow (2\sqrt{x+1}-\sqrt{1-x}+1)(\sqrt{x+1}-\sqrt{1-x})=0$
$+)\sqrt{x+1}-\sqrt{1-x}=0$
$\rightarrow x=0$
$+)2\sqrt{x+1}-\sqrt{1-x}+1=0$
$\rightarrow 2\sqrt{x+1}+1=\sqrt{1-x}$
$\rightarrow (2\sqrt{x+1}+1)^2=1-x$
$\rightarrow 4(x+1)+4\sqrt{x+1}+1=1-x$
$\rightarrow 5(x+1)+4\sqrt{x+1}-1=0$
$\rightarrow (\sqrt{x+1}+1)(5\sqrt{x+1}-1)=0$
$\rightarrow 5\sqrt{x+1}-1=0$
$\rightarrow x=\dfrac{1}{25}-1$
