2 ⋅ 5 2 {\displaystyle 2\cdot 5^{2}}
-- vorläufig vergessen --
7 2 + 1 {\displaystyle 7^{2}+1}
10 ⋅ 10 2 {\displaystyle {10\cdot 10} \over 2}
4 ⋅ 5 ⋅ 125 {\displaystyle {\sqrt {4\cdot 5\cdot 125}}}
∑ i ∈ { 1..10 } ∖ 5 i {\displaystyle \sum _{i\in \{1..10\}\setminus 5}i}