Solution 3.1:3d
From Förberedande kurs i matematik 1
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- | We can multiply | + | We can multiply <math>\sqrt{\tfrac{2}{3}}</math> into the bracket and then write the root expressions together under a common root sign using the rule <math>\sqrt{a\vphantom{b}}\cdot \sqrt{b} = \sqrt{ab}</math>, |
- | <math>\sqrt{\ | + | |
- | into the bracket and then write | + | |
- | <math>\sqrt{a}\ | + | |
+ | {{Displayed math||<math>\sqrt{\frac{2}{3}}\bigl(\sqrt{6}-\sqrt{3}\bigr) = \sqrt{\frac{2}{3}}\cdot\sqrt{6} - \sqrt{\frac{2}{3}}\cdot\sqrt{3} = \sqrt{\frac{2\cdot 6}{3}} - \sqrt{\frac{2\cdot 3}{3}}\,\textrm{.}</math>}} | ||
+ | Because <math>(2\cdot 6)/3 = 2\cdot 2 = 2^2</math> and <math>(2\cdot 3)/3 = 2</math>, we obtain | ||
- | <math>\sqrt{\frac{2}{3}}\ | + | {{Displayed math||<math>\sqrt{\frac{2}{3}}\bigl(\sqrt{6} - \sqrt{3}\bigr) = \sqrt{2^2}-\sqrt{2} = 2-\sqrt{2}\,\textrm{.}</math>}} |
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Current revision
We can multiply \displaystyle \sqrt{\tfrac{2}{3}} into the bracket and then write the root expressions together under a common root sign using the rule \displaystyle \sqrt{a\vphantom{b}}\cdot \sqrt{b} = \sqrt{ab},
\displaystyle \sqrt{\frac{2}{3}}\bigl(\sqrt{6}-\sqrt{3}\bigr) = \sqrt{\frac{2}{3}}\cdot\sqrt{6} - \sqrt{\frac{2}{3}}\cdot\sqrt{3} = \sqrt{\frac{2\cdot 6}{3}} - \sqrt{\frac{2\cdot 3}{3}}\,\textrm{.} |
Because \displaystyle (2\cdot 6)/3 = 2\cdot 2 = 2^2 and \displaystyle (2\cdot 3)/3 = 2, we obtain
\displaystyle \sqrt{\frac{2}{3}}\bigl(\sqrt{6} - \sqrt{3}\bigr) = \sqrt{2^2}-\sqrt{2} = 2-\sqrt{2}\,\textrm{.} |