Lösung 3.1:4c
Aus Online Mathematik Brückenkurs 1
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Each term in the expression can be simplified by breaking down the number under the root sign into its factors, | Each term in the expression can be simplified by breaking down the number under the root sign into its factors, | ||
+ | {{Displayed math||<math>\begin{align} | ||
+ | 50 &= 5\cdot 10 = 5\cdot 5\cdot 2 = 2\cdot 5^{2}\,,\\[5pt] | ||
+ | 20 &= 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5\,,\\[5pt] | ||
+ | 18 &= 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}\,,\\[5pt] | ||
+ | 80 &= 8\cdot 10 = (2\cdot 4)\cdot (2\cdot 5) = (2\cdot 2\cdot 2)\cdot (2\cdot 5) = 2^{4}\cdot 5\,, | ||
+ | \end{align}</math>}} | ||
- | + | and then taking the squares out from under the root sign, | |
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- | and then taking the squares out from under the root sign | + | |
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+ | {{Displayed math||<math>\begin{align} | ||
+ | \sqrt{50} &= \sqrt{2\cdot 5^2} = 5\sqrt{2}\,,\\ | ||
+ | \sqrt{20} &= \sqrt{2^2\cdot 5} = 2\sqrt{5}\,,\\ | ||
+ | \sqrt{18} &= \sqrt{2\cdot 3^2} = 3\sqrt{2}\,,\\ | ||
+ | \sqrt{80} &= \sqrt{2^4\cdot 5} = 2^{2}\sqrt{5} = 4\sqrt{5}\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
All together, we get | All together, we get | ||
- | + | {{Displayed math||<math>\begin{align} | |
- | <math>\begin{align} | + | \sqrt{50} + 4\sqrt{20} - 3\sqrt{18} - 2\sqrt{80} |
- | + | &= 5\sqrt{2} + 4\cdot 2\sqrt{5} - 3\cdot 3\sqrt{2} - 2\cdot 4\sqrt{5}\\[5pt] | |
- | & =5\sqrt{2}+4\ | + | &= 5\sqrt{2} + 8\sqrt{5} - 9\sqrt{2} - 8\sqrt{5}\\[5pt] |
- | & =5\sqrt{2}+8\sqrt{5}-9\sqrt{2}-8\sqrt{5} \\ | + | &= (5-9)\sqrt{2} + (8-8)\sqrt{5} = -4\sqrt{2}\,\textrm{.} |
- | & = | + | \end{align}</math>}} |
- | \end{align}</math> | + |
Version vom 11:05, 30. Sep. 2008
Each term in the expression can be simplified by breaking down the number under the root sign into its factors,
and then taking the squares out from under the root sign,
All together, we get