Lösung 3.1:4c
Aus Online Mathematik Brückenkurs 1
Each term in the expression can be simplified by breaking down the number under the root sign into its factors,
\displaystyle \begin{align}
& 50=5\centerdot 10=5\centerdot 5\centerdot 2=2\centerdot 5^{2} \\
& 20=2\centerdot 10=2\centerdot 2\centerdot 5=2^{2}\centerdot 5 \\
& 18=2\centerdot 9=2\centerdot 3\centerdot 3=2\centerdot 3^{2} \\
& 80=8\centerdot 10=\left( 2\centerdot 4 \right)\centerdot \left( 2\centerdot 5 \right)=\left( 2\centerdot 2\centerdot 2 \right)\centerdot \left( 2\centerdot 5 \right)=2^{4}\centerdot 5 \\
\end{align}
and then taking the squares out from under the root sign.
\displaystyle \begin{align}
& \sqrt{50}=\sqrt{2\centerdot 5^{2}}=5\sqrt{2} \\
& \sqrt{20}=\sqrt{2^{2}\centerdot 5}=2\sqrt{5} \\
& \sqrt{18}=\sqrt{2\centerdot 3^{2}}=3\sqrt{2} \\
& \sqrt{80}=\sqrt{2^{4}\centerdot 5}=2^{2}\sqrt{5}=4\sqrt{5} \\
\end{align}
All together, we get
\displaystyle \begin{align}
& \sqrt{50}+4\sqrt{20}-3\sqrt{18}-2\sqrt{80} \\
& =5\sqrt{2}+4\centerdot 2\sqrt{5}-3\centerdot 3\sqrt{2}-2\centerdot 4\sqrt{5} \\
& =5\sqrt{2}+8\sqrt{5}-9\sqrt{2}-8\sqrt{5} \\
& =\left( 5-9 \right)\sqrt{2}+\left( 8-8 \right)\sqrt{5}=-4\sqrt{2} \\
\end{align}