Với giá trị nào của x thì mỗi căn thức sau cs nghĩa √4-x^2 √x^2-16 √x^2-3 √x^2-2x-3 √x(x+2) √x^2-5x+6

a/ ĐKXĐ: \(4-x^2\ge 0\\↔(2-x)(2+x)\ge 0\\↔\left[\begin{array}{1}\begin{cases}2-x\ge 0\\2+x\ge 0\end{cases}\\\begin{cases}2-x\le 0\\2+x\le 0\end{cases}\end{array}\right.↔\left[\begin{array}{1}\begin{cases}x\le 2\\x\ge -2\end{cases}\\\begin{cases}x\ge 2\\x\le -2\end{cases}\end{array}\right.\\↔-2\le x\le 2\)

b/ ĐKXĐ: \(x^2-16\ge 0\\↔(x-4)(x+4)\ge 0\\↔\left[\begin{array}{1}\begin{cases}x-4\ge 0\\x+4\ge 0\end{cases}\\\begin{cases}x-4\le 0\\x+4\le 0\end{cases}\end{array}\right.↔\left[\begin{array}{1}\begin{cases}x\ge 4\\x\ge -4\end{cases}\\\begin{cases}x\le 4\\x\le -4\end{cases}\end{array}\right.↔\left[\begin{array}{1}x\ge 4\\x\le -4\end{array}\right.\)

c/ ĐKXĐ: \(x^2-3\ge 0\\↔(x-\sqrt 3)(x+\sqrt 3)\ge 0\\↔\left[\begin{array}{1}\begin{cases}x-\sqrt 3\ge 0\\x+\sqrt 3\ge 0\end{cases}\\\begin{cases}x-\sqrt 3\le 0\\x+\sqrt 3\le 0\end{cases}\end{array}\right.↔\left[\begin{array}{1}\begin{cases}x\ge \sqrt 3\\x\ge -\sqrt 3\end{cases}\\\begin{cases}x\le \sqrt 3\\x\le -\sqrt 3\end{cases}\end{array}\right.↔\left[\begin{array}{1}x\ge \sqrt 3\\x\le -\sqrt3\end{array}\right.\)

d/ ĐKXĐ: \(x^2-2x-3\ge 0\\↔x^2-3x+x-3\ge 0\\↔(x^2-3x)+(x-3)\ge 0\\↔x(x-3)+(x-3)\ge 0\\↔(x+1)(x-3)\ge 0\\↔\left[\begin{array}{1}\begin{cases}x+1\ge 0\\x-3\ge 0\end{cases}\\\begin{cases}x+1\le 0\\x-3\le 0\end{cases}\end{array}\right.↔\left[\begin{array}{1}\begin{cases}x\ge -1\\x\ge 3\end{cases}\\\begin{cases}x\le -1\\x\le 3\end{cases}\end{array}\right.↔\left[\begin{array}{1}x\ge 3\\x\le -1\end{array}\right.\)

e/ ĐKXĐ: \(x(x+2)\ge 0\\↔\left[\begin{array}{1}\begin{cases}x\ge 0\\x+2\ge 0\end{cases}\\\begin{cases}x\le 0\\x+2\le 0\end{cases}\end{array}\right.↔\left[\begin{array}{1}\begin{cases}x\ge 0\\x\ge -2\end{cases}\\\begin{cases}x\le 0\\x\le -2\end{cases}\end{array}\right.↔\left[\begin{array}{1}x\ge 0\\x\le -2\end{array}\right.\)

f/ ĐKXĐ: \(x^2-5x+6\ge 0\\↔x^2-2x-3x+6\ge 0\\↔(x^2-2x)-(3x-6)\ge 0\\↔x(x-2)-3(x-2)\ge 0\\↔(x-3)(x-2)\ge 0\\↔\left[\begin{array}{1}\begin{cases}x-3\ge 0\\x-2\ge 0\end{cases}\\\begin{cases}x-3\le 0\\x-2\le 0\end{cases}\end{array}\right.↔\left[\begin{array}{1}\begin{cases}x\ge 3\\x\ge 2\end{cases}\\\begin{cases}x\le 3\\x\le 2\end{cases}\end{array}\right.↔\left[\begin{array}{1}x\ge 3\\x\le 2\end{array}\right.\)