Đk: -1≤x≤1
Đặt $\sqrt{x+1}=a $; $\sqrt{1-x}=b $$ (a;b≥0)$
Khi đó pttt:
$a+2a^2=-b^2+b+3ab$
⇔$2a^2+b^2-3ab+a-b=0$
⇔$(a-b)(2a-b)+(a-b)=0$
⇔\(\left[ \begin{array}{l}a=b\\2a-b=-1\end{array} \right.\)
TH 1: a=b ⇔$\sqrt{x+1}=\sqrt{1-x}$
⇔ x+1=1-x
⇔x=0
TH2: 2a-b=-1 ⇔2$\sqrt{x+1}=\sqrt{1-x}-1$ . ĐK:$\sqrt{1-x}≥1$⇔x≤0⇒-1≤x≤0
⇒4(x+1)=1-x-$2\sqrt{1-x}$+1
⇔$5x+2=-2\sqrt{1-x}$. Đk: 5x+2≤0⇔x≤-2/5⇒-1≤x≤-2/5
⇒$(5x+2)^2=4(1-x)$
⇔x=-24/25 (T/m)
