Решить уравнение х+1=корень из 1-х

[latex]x+1=\sqrt{1-x}\; \; ODZ:\; x+1 \geq 0\; , x\geq -1\\\\\\\\(x+1)^2=1-x\\\\x^2+2x+1=1-x\\\\x^2+3x=0\\\\x(x+3)=0\\\\x_1=0\; ,\; x_2=-3\notin ODZ\\\\Otvet:\; x=0.[/latex]

[latex] \left \{ {{x+1 = \sqrt{1-x} } \atop {x+1 \ \textgreater \ 0}} \right. \left \{ {{x^{2} +2x +1 = 1-x \atop {x\ \textgreater \ -1}} \right. \left \{ {{x^{2} +3x= 0 \atop {x\ \textgreater \ -1}} \right. \left \{ {{x(x+3)= 0 \atop {x\ \textgreater \ -1}} \right. \left \{ {{ \left \{ {{x=0} \atop {x=-3 }} \right. \atop {x\ \textgreater \ -1}} \right. x = 0[/latex] Ответ: 0