Lösung 3.1:4d
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Version vom 14:22, 22. Okt. 2008
We start by factorizing the numbers under the root sign,
| \displaystyle \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} |
Now, we can take the squares out from under the root signs,
| \displaystyle \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} |
and then simplify the whole expression
| \displaystyle \begin{align}
\sqrt{48} + \sqrt{12} + \sqrt{3} - \sqrt{75} &= 4\sqrt{3} + 2\sqrt{3} + \sqrt{3} - 5\sqrt{3}\\[5pt] &= (4+2+1-5)\sqrt{3}\\[5pt] &= 2\sqrt{3}\,\textrm{.} \end{align} |
