Đáp án:
$D = \frac{3x-2\sqrt[]{x}+2}{2\sqrt[]{x}-2} + \frac{2\sqrt[]{x}+1}{2-2\sqrt[]{x}}$ $( x ≥ 0 , x \ne 1 )$
⇔ $D = \frac{3x-2\sqrt[]{x}+2}{2\sqrt[]{x}-2} - \frac{2\sqrt[]{x}+1}{2\sqrt[]{x}-2}$
⇔ $D = \frac{3x-2\sqrt[]{x}+2-2\sqrt[]{x}-1}{2\sqrt[]{x}-2}$
⇔ $D = \frac{3x-4\sqrt[]{x}+1}{2(\sqrt[]{x}-1)}$
⇔ $D = \frac{(\sqrt[]{x}-1)(3\sqrt[]{x}-1)}{2(\sqrt[]{x}-1)}$
⇔ $D = \frac{3\sqrt[]{x}-1}{2}$
$F = \frac{2x+3\sqrt[]{x}-1}{\sqrt[]{x}-1} - \frac{3\sqrt[]{x}+1}{\sqrt[]{x}-1}$ $( x ≥ 0 , x \ne 1 )$
⇔ $F = \frac{2x+3\sqrt[]{x}-1-3\sqrt[]{x}-1}{\sqrt[]{x}-1}$
⇔ $F = \frac{2x-2}{\sqrt[]{x}-1}$
⇔ $F = \frac{2(x-1)}{\sqrt[]{x}-1}$
⇔ $F = \frac{2(\sqrt[]{x}-1)(\sqrt[]{x}+1)}{\sqrt[]{x}-1}$
⇔ $F = 2( \sqrt[]{x} + 1 )$
