Подкоренное выражение корня чётной степени должно быть неотрицательным , то есть  ≥ 0 :  

[tex]\displaystyle\bf\\1)\\\\\sqrt{x+4} +\sqrt{5-x} \\\\\\\left \{ {{x+4\geq 0} \atop {5-x\geq 0}} \right. \\\\\\\left \{ {{x\geq -4} \atop {x\leq 5}} \right. \\\\\\Otvet \ : \ x\in\Big[-4 \ ; \ 5\Big]\\\\2)\\\\\sqrt{10-2x} +\sqrt{49+7x} \\\\\\\left \{ {{10-2x\geq 0} \atop {49+7x\geq 0}} \right. \\\\\\\left \{ {{-2x\geq -10} \atop {7x\geq -49}} \right. \\\\\\\left \{ {{x\leq 5} \atop {x\geq -7}} \right. \\\\\\Otvet \ : \ x\in\Big[-7 \ ; \ 5\Big][/tex]

[tex]\displaystyle\bf\\ 3)\\\\\sqrt{x^{2} -9} +\sqrt{x-2} \\\\\\\left \{ {{x^{2} -9\geq 0} \atop {x-2\geq 0}} \right. \\\\\\\left \{ {{(x-3)\cdot(x+3)\geq 0} \atop {x\geq 2}} \right. \\\\\\\left \{ {{x\in(-\infty \ ; \ -3]\cup[3 \ ; \ +\infty)} \atop {x\geq 2}} \right. \\\\\\Otvet \ : \ x\in \ \Big [3 \ ; \ +\infty\Big)[/tex]

[tex]\displaystyle\bf\\4)\\\\\sqrt{4+x} +\sqrt{16-x^{2} } \\\\\\\left \{ {{16-x^{2} \geq 0} \atop {4+x\geq 0}} \right. \\\\\\\left \{ {{(4-x)\cdot(4+x)\geq 0} \atop {x\geq -4}} \right. \\\\\\\left \{ {{x\in[-4 \ ; \ 4]} \atop {x\geq -4}} \right. \\\\\\Otvet \ : \ x\in \ \Big [-4 \ ; \ 4\Big][/tex]