ĐK : \(-1\le x\le1\)
+ Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+1}\ge0\\b=\sqrt{1-x}\ge0\end{matrix}\right.\) thì pt đã cho trở thành :
\(a+2a^2=-b^2+b+3ab\)
\(\Rightarrow a+2a^2+b^2-b-3ab=0\)
\(\Rightarrow a\left(2a-b+1\right)-b\left(2a-b+1\right)=0\)
\(\Rightarrow\left(a-b\right)\left(2a-b+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\2a=b-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{matrix}\right.\)
+ TH1 : \(\sqrt{x+1}=\sqrt{1-x}\Leftrightarrow x+1=1-x\Leftrightarrow x=0\) ( TM )
+ TH2 : \(2\sqrt{x+1}+1=\sqrt{1-x}\)
\(\Leftrightarrow4x+5+4\sqrt{x+1}=1-x\)
\(\Leftrightarrow4\sqrt{x+1}=-5x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}-5x-4\ge0\\16\left(x+1\right)=\left(-5x-4\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\frac{4}{5}\\16x+16=25x^2+40x+16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le-\frac{4}{5}\\25x^2+24x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le-\frac{4}{5}\\x\left(25x+24\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le-\frac{4}{5}\\\left[{}\begin{matrix}x=0\left(KTM\right)\\x=-\frac{24}{25}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)
