Решите уравнение: корень из (x+1)+корень из (4-x)=3.

\[\sqrt{x + 1} + \sqrt{4 - x} = 3\]

\[ОДЗ:x \geq - 1;x \leq 4;\]

\[- 1 \leq x \leq 4.\]

\[\left( \sqrt{x + 1} \right)^{2} = \left( 3 - \sqrt{4 - x} \right)^{2}\]

\[x + 1 = 9 - 6\sqrt{4 - x} + 4 - x\]

\[6\sqrt{4 - x} = 12 - 2x\]

\[36(4 - x) = (12 - 2x)^{2}\]

\[144 - 36x = 144 - 48x + 4x^{2}\]

\[4x^{2} - 12x = 0\]

\[4x(x - 3) = 0\]

\[x = 0;\ \ \ x = 3.\]

\[Ответ:x = 0;\ \ x = 3.\]