Lösung 3.1:4d
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K |
|||
Zeile 1: | Zeile 1: | ||
We start by factorizing the numbers under the root sign, | We start by factorizing the numbers under the root sign, | ||
- | + | {{Displayed math||<math>\begin{align} | |
- | <math>\begin{align} | + | 48 &= 2\cdot 24 = 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 6 = 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{4}\cdot 3\,,\\ |
- | + | 12 &= 2\cdot 6 = 2\cdot 2\cdot 3 = 2^{2}\cdot 3\,,\\ | |
- | + | 3 &= 3\,,\\ | |
- | + | 75 &= 3\cdot 25 = 3\cdot 5\cdot 5 = 3\cdot 5^{2}\,\textrm{.} | |
- | + | \end{align}</math>}} | |
- | \end{align}</math> | + | |
Now, we can take the squares out from under the root signs, | Now, we can take the squares out from under the root signs, | ||
+ | {{Displayed math||<math>\begin{align} | ||
+ | \sqrt{48} &= \sqrt{2^4\cdot 3} = 2^2\sqrt{3} = 4\sqrt{3}\,,\\[5pt] | ||
+ | \sqrt{12} &= \sqrt{2^2\cdot 3} = 2\sqrt{3},\\[5pt] | ||
+ | \sqrt{3} &= \sqrt{3}\,,\\[5pt] | ||
+ | \sqrt{75} &= \sqrt{3\cdot 5^{2}} = 5\sqrt{3}\,, | ||
+ | \end{align}</math>}} | ||
- | + | and then simplify the whole expression | |
- | + | ||
- | + | ||
- | + | ||
- | + | ||
- | + | ||
- | + | ||
- | + | ||
- | and then simplify the whole expression | + | |
- | + | ||
- | <math>\begin{align} | + | {{Displayed math||<math>\begin{align} |
- | + | \sqrt{48} + \sqrt{12} + \sqrt{3} - \sqrt{75} | |
- | & = | + | &= 4\sqrt{3} + 2\sqrt{3} + \sqrt{3} - 5\sqrt{3}\\[5pt] |
- | \end{align}</math> | + | &= (4+2+1-5)\sqrt{3}\\[5pt] |
+ | &= 2\sqrt{3}\,\textrm{.} | ||
+ | \end{align}</math>}} |
Version vom 11:13, 30. Sep. 2008
We start by factorizing the numbers under the root sign,
Now, we can take the squares out from under the root signs,
and then simplify the whole expression