96/VB02. problem / Math - Solved Math Problems

MR-96 / 100a. problem

Hozza egyszerűbb alakra a következő kifejezést:

$\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$= \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \cdot \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

$= \frac{{(\sqrt{3} + \sqrt{2})}^{2}}{{(\sqrt{3})}^{2} - {(\sqrt{2})}^{2}}$

$= \frac{{(\sqrt{3} + \sqrt{2})}^{2}}{3 - 2}$

$= \frac{{(\sqrt{3})}^{2} + 2 \sqrt{3} \sqrt{2} + {(\sqrt{2})}^{2}}{1}$

$= 3 + 2 \sqrt{3} \sqrt{2} + 2$

$= 5 + 2 \sqrt{6}$

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