Solution 3.1:6a
From Förberedande kurs i matematik 1
(Difference between revisions)
m |
|||
| Line 1: | Line 1: | ||
| - | We use the standard method and augment the fraction with the conjugate of the denominator | + | We use the standard method and augment the fraction with the conjugate of the denominator <math>\sqrt{5}+2</math>. Then the formula for the difference of two squares gives |
| - | <math>\sqrt{5}+2</math>. Then the | + | |
| - | + | {{Displayed math||<math>\begin{align} | |
| - | <math>\begin{align} | + | \frac{\sqrt{2}+3}{\sqrt{5}-2} |
| - | + | &= \frac{\sqrt{2}+3}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2}\\[5pt] | |
| - | & =\frac{\sqrt{2}\ | + | &= \frac{(\sqrt{2}+3)(\sqrt{5}+2)}{(\sqrt{5})^{2}-2^{2}}\\[5pt] |
| - | & =6+2\sqrt{2}+3\sqrt{5}+10 \\ | + | &= \frac{\sqrt{2}\cdot\sqrt{5}+\sqrt{2}\cdot 2+3\cdot \sqrt{5}+3\cdot 2}{5-4}\\[5pt] |
| - | \end{align}</math> | + | &= \sqrt{2\cdot 5} + 2\sqrt{2} + 3\sqrt{5} + 6\\[5pt] |
| + | &= 6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
We use the standard method and augment the fraction with the conjugate of the denominator \displaystyle \sqrt{5}+2. Then the formula for the difference of two squares gives
| \displaystyle \begin{align}
\frac{\sqrt{2}+3}{\sqrt{5}-2} &= \frac{\sqrt{2}+3}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2}\\[5pt] &= \frac{(\sqrt{2}+3)(\sqrt{5}+2)}{(\sqrt{5})^{2}-2^{2}}\\[5pt] &= \frac{\sqrt{2}\cdot\sqrt{5}+\sqrt{2}\cdot 2+3\cdot \sqrt{5}+3\cdot 2}{5-4}\\[5pt] &= \sqrt{2\cdot 5} + 2\sqrt{2} + 3\sqrt{5} + 6\\[5pt] &= 6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}\,\textrm{.} \end{align} |
