Đáp án:
$a)P=\dfrac{x^2-2x-1}{x-1}\\ b)\left[\begin{array}{l} x \ge 1+\sqrt{2} \\ 1-\sqrt{2}\le x <1\end{array} \right. $
Giải thích các bước giải:
$P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{x^2-3x}{x-\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-3\sqrt{x}+2}\right)\\ =\left(\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-2)}+\dfrac{x^2-3x}{(\sqrt{x}+1)(\sqrt{x}-2)}\right):\left(\dfrac{\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}-1)(\sqrt{x}-2)}+\dfrac{1}{(\sqrt{x}-1)(\sqrt{x}-2)}\right)\\ =\dfrac{x-1+x^2-3x}{(\sqrt{x}+1)(\sqrt{x}-2)}:\dfrac{x-2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}-2)}\\ =\dfrac{x^2-2x-1}{(\sqrt{x}+1)(\sqrt{x}-2)}.\dfrac{(\sqrt{x}-1)(\sqrt{x}-2)}{(\sqrt{x}-1)^2}\\ =\dfrac{x^2-2x-1}{(\sqrt{x}-1)(\sqrt{x}+1)}\\ =\dfrac{x^2-2x-1}{x-1}\\ b)P>0 \Leftrightarrow \dfrac{x^2-2x-1}{x-1}>0\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x^2-2x-1 \ge 0\\ x-1 >0\end{array} \right.\\ \left\{\begin{array}{l} x^2-2x-1 \le 0\\ x-1 <0 \end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} \left[\begin{array}{l} x \ge 1+\sqrt{2} \\x \le 1-\sqrt{2} \end{array} \right.\\ x>1\end{array} \right.\\ \left\{\begin{array}{l} 1-\sqrt{2}\le x\le 1+\sqrt{2}\\ x<1 \end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x \ge 1+\sqrt{2} \\ 1-\sqrt{2}\le x <1\end{array} \right.$
