Lösung 1.1:2b

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Först identifierar vi det innersta parentesuttrycket och beräknar detta
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When calculating, the innermost part of the expression should always be calculated first.
:<math>3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5) = 3-((\bbox[#FFEEAA;,1.5pt]{\,3\,}+6)-5)</math>.
:<math>3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5) = 3-((\bbox[#FFEEAA;,1.5pt]{\,3\,}+6)-5)</math>.
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I det uttryck som nu uppstår ringar vi in den innersta parentesen och beräknar den
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In the resulting expression we again pick out and calculate the innermost part
:<math>\phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-(\firstcbox{#FFEEAA;}{\,(3+6)\,}{9}-5)</math>
:<math>\phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-(\firstcbox{#FFEEAA;}{\,(3+6)\,}{9}-5)</math>
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:<math>\phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-(\secondcbox{#FFEEAA;}{\,(3+6)\,}{9}-5)</math>.
:<math>\phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-(\secondcbox{#FFEEAA;}{\,(3+6)\,}{9}-5)</math>.
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Efter detta har vi bara en parentes kvar att räkna ut
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Now there is just one bracket to calculate
:<math>\phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-\firstcbox{#FFEEAA;}{\,(9-5)\,}{4}</math>
:<math>\phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-\firstcbox{#FFEEAA;}{\,(9-5)\,}{4}</math>
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:<math>\phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-\secondcbox{#FFEEAA;}{\,(9-5)\,}{4}</math>
:<math>\phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-\secondcbox{#FFEEAA;}{\,(9-5)\,}{4}</math>
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vilket till slut ger ett uttryck utan parenteser som vi direkt kan räkna ut
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and now there is only an expression without brackets left to calculate
:<math>\phantom{3-((\firstcbox{#FFEEAA;}{\,(7-4)\,}{3}+6)-5)}{} = -1</math>.
:<math>\phantom{3-((\firstcbox{#FFEEAA;}{\,(7-4)\,}{3}+6)-5)}{} = -1</math>.
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Version vom 13:10, 13. Sep. 2008

When calculating, the innermost part of the expression should always be calculated first.

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

In the resulting expression we again pick out and calculate the innermost part

\displaystyle \phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-(\firstcbox{#FFEEAA;}{\,(3+6)\,}{9}-5)
\displaystyle \phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-(\secondcbox{#FFEEAA;}{\,(3+6)\,}{9}-5).

Now there is just one bracket to calculate

\displaystyle \phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-\firstcbox{#FFEEAA;}{\,(9-5)\,}{4}
\displaystyle \phantom{3-((\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}+6)-5)}{} = 3-\secondcbox{#FFEEAA;}{\,(9-5)\,}{4}

and now there is only an expression without brackets left to calculate

\displaystyle \phantom{3-((\firstcbox{#FFEEAA;}{\,(7-4)\,}{3}+6)-5)}{} = -1.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.1:2b“