Đáp án:
6
Giải thích các bước giải:
$d) \left ( \frac{\sqrt{x}+1}{x-4}-\frac{\sqrt{x}-1}{x+4\sqrt{x}+4} \right ).\frac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\\
= \left ( \frac{\sqrt{x}+1}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{\sqrt{x}-1}{(\sqrt{x}+2)^2} \right ).\frac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\\
=\left ( \frac{(\sqrt{x}+2)(\sqrt{x}+1)}{(\sqrt{x}-2)(\sqrt{x}+2)^2}-\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{(\sqrt{x}-2)(\sqrt{x}+2)^2} \right ).\frac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\\
=\frac{(\sqrt{x}+2)(\sqrt{x}+1)-(\sqrt{x}-2)(\sqrt{x}-1)}{(\sqrt{x}-2)(\sqrt{x}+2)^2}.\frac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\\
=\frac{x+\sqrt{x}+2\sqrt{x}+2-(x-\sqrt{x}-2\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)^2}.\frac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\\=\frac{x+\sqrt{x}+2\sqrt{x}+2-x+\sqrt{x}+2\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)^2}.\frac{x(\sqrt{x}+2)-4(\sqrt{x}+2)}{\sqrt{x}}\\
=\frac{6\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)^2}.\frac{(\sqrt{x}+2)(x-4)}{\sqrt{x}}\\
=\frac{6}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{(x-4)}{1}\\
=\frac{6}{x-4}.\frac{(x-4)}{1}\\
=6$
