Lösung 1.1:1d

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:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 1</math>.
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 1</math>.
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Version vom 06:41, 19. Sep. 2008

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.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.1:1d“