Đáp án:

$5)\\a) \left[\begin{array}{l} x \ge 4\\ x < 0\end{array} \right.\\ b)\sqrt{x^2-4x}\\ c) \Leftrightarrow \left[\begin{array}{l} 4 \le x <5\\ -1<x<0\end{array} \right.\\ 6)\\ a)A=-2\sqrt{x}\\ b) 0< x <9, x \ne 1$

Giải thích các bước giải:

$5)A=\dfrac{x+\sqrt{x^2-4x}}{x-\sqrt{x^2-4x}}-\dfrac{x-\sqrt{x^2-4x}}{x+\sqrt{x^2-4x}}\\ \text{ĐKXĐ}: \left\{\begin{array}{l} x^2-4x \ge 0\\ x-\sqrt{x^2-4x} \ne 0\\ x+\sqrt{x^2-4x} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x(x-4) \ge 0\\ x \ne \sqrt{x^2-4x}\\ \sqrt{x^2-4x} \ne -x \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 4\\ x \le 0\end{array} \right.\\ x^2 \ne x^2-4x\\ x \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 4\\ x \le 0\end{array} \right.\\ -4x \ne 0\\ x \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 4\\ x \le 0\end{array} \right.\\x \ne 0 \end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x \ge 4\\ x < 0\end{array} \right.\\ b)\dfrac{x+\sqrt{x^2-4x}}{x-\sqrt{x^2-4x}}-\dfrac{x-\sqrt{x^2-4x}}{x+\sqrt{x^2-4x}}\\ =\dfrac{(x+\sqrt{x^2-4x})^2}{(x-\sqrt{x^2-4x})(x+\sqrt{x^2-4x})}-\dfrac{(x-\sqrt{x^2-4x})^2}\\{(x+\sqrt{x^2-4x})(x-\sqrt{x^2-4x})}\\ =\dfrac{(x+\sqrt{x^2-4x})^2-(x-\sqrt{x^2-4x})^2}{(x+\sqrt{x^2-4x})(x-\sqrt{x^2-4x})}\\ =\dfrac{4x\sqrt{x^2-4x}}{x^2-(x^2-4x)}\\ =\dfrac{4x\sqrt{x^2-4x}}{4x}\\ =\sqrt{x^2-4x}\\ c)A<\sqrt{5}\\ \Leftrightarrow  \sqrt{x^2-4x} <\sqrt{5} \\ \Rightarrow  0 \le x^2-4x <5\\ \Rightarrow \left\{\begin{array}{l} x^2-4x \ge 0\\ x^2-4x <5\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 4\\ x < 0\end{array} \right.\\ x^2-4x -5<0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 4\\ x < 0\end{array} \right.\\ x^2+x-5x -5<0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 4\\ x < 0\end{array} \right.\\ x(x+1)-5(x +1)<0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 4\\ x < 0\end{array} \right.\\ (x-5)(x+1)<0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 4\\ x < 0\end{array} \right.\\ \left[\begin{array}{l} \left\{\begin{array}{l} x-5<0 \\ x+1>0\end{array} \right.\\ \left\{\begin{array}{l} x-5>0 \\ x+1<0\end{array} \right.\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 4\\ x < 0\end{array} \right.\\ \left[\begin{array}{l} \left\{\begin{array}{l} x<5 \\ x>-1\end{array} \right.\\ \left\{\begin{array}{l} x>5 \\ x<-1\end{array} \right.(\text{Vô lí})\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 4\\ x < 0\end{array} \right.\\ -1 < x<5\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} 4 \le x <5\\ -1<x<0\end{array} \right.\\ 6)\\ A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\\ \text{ĐKXĐ:}\left\{\begin{array}{l} x \ge 0 \\ 2\sqrt{x} \ne 0 \\ \sqrt{x}+1 \ne 0\\ \sqrt{x}-1 \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ \sqrt{x} \ne 0 \\ \sqrt{x} \ne -1\\ \sqrt{x} \ne 1 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ x \ne 0 \\ x\ne 1 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x > 0 \\ x\ne 1 \end{array} \right.\\ a)A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\\ =\left(\dfrac{x}{2\sqrt{x}}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{(x-\sqrt{x})(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}-\dfrac{(x+\sqrt{x})(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\right)\\ =\dfrac{x-1}{2\sqrt{x}}.\dfrac{(x-\sqrt{x})(\sqrt{x}-1)-(x+\sqrt{x})(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\ =\dfrac{x-1}{2\sqrt{x}}.\dfrac{-4x}{x-1}\\ =-2\sqrt{x}\\ b)A>-6\\ \Leftrightarrow -2\sqrt{x}  >-6\\ \Leftrightarrow \sqrt{x}  < 3\\ \Leftrightarrow  0\le x <9\\ \text{Kết hợp điều kiện} \Rightarrow 0< x <9, x \ne 1$