Lösung 1.1:1d

Just as in exercise c, we calculate the innermost bracket

\displaystyle 3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5 = 3-(7-\bbox[#FFEEAA;,1.5pt]{\,10\,})-5

and work our way successively outwards,

\displaystyle \phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 3-\firstcbox{#FFEEAA;}{\,(7-10)\,}{(-3)}-5

\displaystyle \phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 3-\secondcbox{#FFEEAA;}{\,(7-10)\,}{(-3)}-5\,.

All that remains is to combine the terms from left to right

\displaystyle \phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = \firstcbox{#FFEEAA;}{\,3-(-3)\,}{6}-5

\displaystyle \phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = \secondcbox{#FFEEAA;}{\,3-(-3)\,}{3+3}-5

\displaystyle \phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = \secondcbox{#FFEEAA;}{\,3-(-3)\,}{6}-5

\displaystyle \phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 1.