🧮 تمارين الأعداد الجذرية 🧮

$$ \begin{align} (\sqrt{2}-1)^2 &= 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} \\ \frac{5 - (3 - 2\sqrt{2})}{4} &= \frac{5 - 3 + 2\sqrt{2}}{4} = \frac{2 + 2\sqrt{2}}{4} = \frac{1 + \sqrt{2}}{2} \\ \frac{7}{4} - \frac{\sqrt{6}}{2} &= \frac{7 - 2\sqrt{6}}{4} \\ a &= \sqrt{\frac{7 - 2\sqrt{6}}{4}} + \frac{1 + \sqrt{2}}{2} \\ &= \frac{\sqrt{7 - 2\sqrt{6}}}{2} + \frac{1 + \sqrt{2}}{2} \\ &= \frac{\sqrt{(\sqrt{6}-1)^2} + 1 + \sqrt{2}}{2} \\ &= \frac{\sqrt{6}-1 + 1 + \sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{2} \end{align} $$

$$ \begin{align} \frac{\sqrt{2}}{\sqrt{3} - 1} &= \frac{\sqrt{2}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\ &= \frac{\sqrt{6} + \sqrt{2}}{3 - 1} = \frac{\sqrt{6} + \sqrt{2}}{2} \\ b &= \sqrt{6} - \frac{\sqrt{6} + \sqrt{2}}{2} \\ &= \frac{2\sqrt{6} - (\sqrt{6} + \sqrt{2})}{2} \\ &= \frac{2\sqrt{6} - \sqrt{6} - \sqrt{2}}{2} \\ &= \frac{\sqrt{6} - \sqrt{2}}{2} \end{align} $$

$$ \begin{align} a - b &= \frac{\sqrt{6} + \sqrt{2}}{2} - \frac{\sqrt{6} - \sqrt{2}}{2} \\ &= \frac{\sqrt{6} + \sqrt{2} - \sqrt{6} + \sqrt{2}}{2} \\ &= \frac{2\sqrt{2}}{2} = \sqrt{2} \\ \text{بما أن } \sqrt{2} > 0 \text{، فإن } a > b \end{align} $$

$$ \begin{align} \text{لدينا } b &= \frac{\sqrt{6} - \sqrt{2}}{2} \\ \text{نعلم أن } 4 < 6 < 9 &\Rightarrow \sqrt{4} < \sqrt{6} < \sqrt{9} \Rightarrow 2 < \sqrt{6} < 3 \\ \text{و } 1 < 2 < 4 &\Rightarrow \sqrt{1} < \sqrt{2} < \sqrt{4} \Rightarrow 1 < \sqrt{2} < 2 \\ \text{إذن: } \sqrt{6} - \sqrt{2} > 2 - 2 = 0 &\Rightarrow b > 0 \\ \text{و: } \sqrt{6} - \sqrt{2} < 3 - 1 = 2 &\Rightarrow \frac{\sqrt{6} - \sqrt{2}}{2} < \frac{2}{2} = 1 \\ \text{ومنه نستنتج أن } 0 < b < 1 \end{align} $$

$$ \begin{align} \frac{1}{1 - x} - (x + 1) &= \frac{1 - (1 - x)(1 + x)}{1 - x} \\ &= \frac{1 - (1 - x^2)}{1 - x} \\ &= \frac{1 - 1 + x^2}{1 - x} = \frac{x^2}{1 - x} \\ \text{بما أن } 0 < x < 1 \text{، فإن } x^2 > 0 \text{ و } 1-x > 0 \\ \text{إذن } \frac{x^2}{1 - x} > 0 \\ \text{وبالتالي } \frac{1}{1 - x} - (x + 1) > 0 \\ \Rightarrow x + 1 < \frac{1}{1 - x} \end{align} $$

$$ \begin{align} \text{لنضع } x = \frac{\sqrt{6}-\sqrt{2}}{2} \text{ (وهو العدد b من تمرين سابق)} \\ \text{لقد أثبتنا أن } 0 < x < 1 \\ \text{وبالتالي، يمكننا تطبيق المتراجحة السابقة: } \\ x+1 &< \frac{1}{1-x} \\ \frac{\sqrt{6}-\sqrt{2}}{2} + 1 &< \frac{1}{1 - \frac{\sqrt{6}-\sqrt{2}}{2}} \\ \frac{\sqrt{6}-\sqrt{2}+2}{2} &< \frac{1}{\frac{2 - (\sqrt{6}-\sqrt{2})}{2}} \\ \frac{\sqrt{6}+2-\sqrt{2}}{2} &< \frac{2}{2+\sqrt{2}-\sqrt{6}} \\ \text{بضرب الطرفين في 2 نحصل على:} \\ \sqrt{6}+2-\sqrt{2} &< \frac{4}{2+\sqrt{2}-\sqrt{6}} \end{align} $$