Đáp án:

$A=\dfrac{-3+4\sqrt{x}-2\sqrt{x-5}}{x\sqrt{x}+1} \ \ \ \ ĐKXĐ: x\ge 0\\ B=\dfrac{-5\sqrt{x}+2}{\sqrt{x}-1} \ \ \ \ ĐKXĐ:  \left\{\begin{array}{l} x \ge 0 \\ x \ne 1\end{array} \right.\\ C=-\dfrac{\sqrt{x}+6}{\sqrt{x}-2} \ \ \ \ ĐKXĐ: \left\{\begin{array}{l} x \ge 0 \\ x \ne 4 \end{array} \right.\\ E=\dfrac{1}{\sqrt{x}-2} \ \ \ \ ĐKXĐ:\left\{\begin{array}{l} x \ge 0 \\x \ne 1 \\ x \ne 4\end{array} \right.$

Giải thích các bước giải:

$A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-2}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}+2\sqrt{x-5}}{x\sqrt{x}+1}\\ ĐKXĐ:\left\{\begin{array}{l} x \ge 0 \\ \sqrt{x}+1 \ne 0 \\x-\sqrt{x}+1 \ne 0 \\ x\sqrt{x}+1 \ne 0\end{array} \right.\\ \Leftrightarrow x\ge 0\\ A=\dfrac{(2\sqrt{x}-1)(x-\sqrt{x}+1)}{(\sqrt{x}+1)(x-\sqrt{x}+1)}+\dfrac{(3\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}+1)(x-\sqrt{x}+1)}-\dfrac{2x\sqrt{x}+2\sqrt{x-5}}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\\ =\dfrac{(2\sqrt{x}-1)(x-\sqrt{x}+1)+(3\sqrt{x}-2)(\sqrt{x}+1)-(2x\sqrt{x}+2\sqrt{x-5})}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\\ =\dfrac{-3+4\sqrt{x}-2\sqrt{x-5}}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\\ =\dfrac{-3+4\sqrt{x}-2\sqrt{x-5}}{x\sqrt{x}+1}\\ B=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\\ ĐKXĐ:\left\{\begin{array}{l} x \ge 0 \\ x+2\sqrt{x}-3 \ne 0 \\\sqrt{x}-1 \ne 0 \\ \sqrt{x}+3 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ (\sqrt{x}+3)(\sqrt{x}-1) \ne 0 \\\sqrt{x}-1 \ne 0 \\ \sqrt{x}+3 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ x \ne 1\end{array} \right.\\ B=\dfrac{15\sqrt{x}-11}{(\sqrt{x}+3)(\sqrt{x}-1)}-\dfrac{(\sqrt{x}+3)(3\sqrt{x}-2)}{(\sqrt{x}+3)(\sqrt{x}-1)}-\dfrac{(\sqrt{x}-1)(2\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-1)}\\ =\dfrac{15\sqrt{x}-11-(\sqrt{x}+3)(3\sqrt{x}-2)-(\sqrt{x}-1)(2\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-1)}\\ =\dfrac{-5x+7\sqrt{x}-2}{(\sqrt{x}+3)(\sqrt{x}-1)}\\ =\dfrac{(-5\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-1)}\\ =\dfrac{-5\sqrt{x}+2}{\sqrt{x}-1}\\ C=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\\ ĐKXĐ:\left\{\begin{array}{l} x \ge 0 \\ \sqrt{x}+1 \ne 0 \\ \sqrt{x}-2 \ne 0 \\ x-\sqrt{x}-2 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ \sqrt{x}+1 \ne 0 \\ \sqrt{x}-2 \ne 0 \\ (\sqrt{x}-2)(\sqrt{x}+1) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ x \ne 4 \end{array} \right.\\ C=\dfrac{(\sqrt{x}-1) (\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+1) }-\dfrac{(\sqrt{x}+3)(\sqrt{x}+1)}{(\sqrt{x}-2)(\sqrt{x}+1) }-\dfrac{x+5}{(\sqrt{x}-2)(\sqrt{x}+1) }\\ =\dfrac{(\sqrt{x}-1) (\sqrt{x}-2)-(\sqrt{x}+3)(\sqrt{x}+1)-(x+5)}{(\sqrt{x}-2)(\sqrt{x}+1) }\\ =\dfrac{-x-7\sqrt{x}-6}{(\sqrt{x}-2)(\sqrt{x}+1) }\\ =\dfrac{-(\sqrt{x}+1)(\sqrt{x}+6)}{(\sqrt{x}-2)(\sqrt{x}+1) }\\ =-\dfrac{\sqrt{x}+6}{\sqrt{x}-2}\\ E=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\\ ĐKXĐ:\left\{\begin{array}{l} x \ge 0 \\ \sqrt{x}-2 \ne 0 \\ \sqrt{x}-1 \ne 0 \\ x-3\sqrt{x}+2 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ \sqrt{x}-2 \ne 0 \\ \sqrt{x}-1 \ne 0 \\ (\sqrt{x}-2)(\sqrt{x}-1) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\x \ne 1 \\ x \ne 4\end{array} \right.\\ E=\dfrac{(\sqrt{x}-3)(\sqrt{x}-1 )}{(\sqrt{x}-2)(\sqrt{x}-1) }-\dfrac{(2\sqrt{x}-1)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-1) }+\dfrac{x-2}{(\sqrt{x}-2)(\sqrt{x}-1) }\\ =\dfrac{(\sqrt{x}-3)(\sqrt{x}-1 )-(2\sqrt{x}-1)(\sqrt{x}-2)+x-2}{(\sqrt{x}-2)(\sqrt{x}-1) }\\ =\dfrac{\sqrt{x}-1}{(\sqrt{x}-2)(\sqrt{x}-1) }\\ =\dfrac{1}{\sqrt{x}-2}$