Lösung 3.1:3c
Aus Online Mathematik Brückenkurs 1
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- | We start by looking at | + | We start by looking at one part of the expression <math>\sqrt{16}</math>. This subexpression can be simplified since <math>16 = 4\cdot 4 = 4^{2}</math> which gives that <math>\sqrt{16} = \sqrt{4^{2}} = 4</math> and the whole expression becomes |
- | <math>\sqrt{16}</math>. This | + | |
- | <math>16=4\ | + | |
- | which gives that | + | |
- | <math>\sqrt{16}=\sqrt{4^{2}}=4</math> | + | |
- | and the whole expression becomes | + | |
+ | {{Displayed math||<math>\sqrt{16+\sqrt{16}} = \sqrt{16+4} = \sqrt{20}\,\textrm{.}</math>}} | ||
- | <math> | + | Can <math>\sqrt{20}</math> be simplified? In order to answer this, we split 20 up into integer factors, |
+ | {{Displayed math||<math>20 = 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5</math>}} | ||
- | + | and see that 20 contains the square <math>2^2</math> as a factor and can therefore be taken outside the root sign, | |
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- | + | {{Displayed math||<math>\sqrt{20} = \sqrt{2^{2}\centerdot 5} = 2\sqrt{5}\,\textrm{.}</math>}} | |
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- | <math>\sqrt{20}=\sqrt{2^{2}\centerdot 5 | + |
Version vom 10:24, 30. Sep. 2008
We start by looking at one part of the expression \displaystyle \sqrt{16}. This subexpression can be simplified since \displaystyle 16 = 4\cdot 4 = 4^{2} which gives that \displaystyle \sqrt{16} = \sqrt{4^{2}} = 4 and the whole expression becomes
Can \displaystyle \sqrt{20} be simplified? In order to answer this, we split 20 up into integer factors,
and see that 20 contains the square \displaystyle 2^2 as a factor and can therefore be taken outside the root sign,