Lösung 1.1:1c

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Parenteserna talar om i vilken ordning uttrycket ska räknas ut och vi börjar med att räkna ut den innersta parentesen
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The brackets indicate in which order the expression is to be calculated, and we start by
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calculating the inner bracket.
:<math>3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5) = 3-(7-\bbox[#FFEEAA;,1.5pt]{\,10\,}-5)</math>.
:<math>3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5) = 3-(7-\bbox[#FFEEAA;,1.5pt]{\,10\,}-5)</math>.
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Sedan räknar vi ut nästa parentes
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Then, we calculate the next bracket.
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=3-\firstcbox{#FFEEAA;}{\,(7-10-5)\,}{(-8)}</math>
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=3-\firstcbox{#FFEEAA;}{\,(7-10-5)\,}{(-8)}</math>
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:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=3-\secondcbox{#FFEEAA;}{\,(7-10-5)\,}{(-8)}</math>
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=3-\secondcbox{#FFEEAA;}{\,(7-10-5)\,}{(-8)}</math>
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och när vi tar bort parentesen i "<math>{}-(-8)</math>" får vi "<math>{}+8</math>",
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and when we remove the brackets in "<math>{}-(-8)</math>" we get "<math>{}+8</math>",
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=3+8</math>
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=3+8</math>
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:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=11.</math>
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=11.</math>
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Version vom 12:37, 13. Sep. 2008

The brackets indicate in which order the expression is to be calculated, and we start by calculating the inner bracket.

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

Then, we calculate the next bracket.

\displaystyle \phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=3-\firstcbox{#FFEEAA;}{\,(7-10-5)\,}{(-8)}
\displaystyle \phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=3-\secondcbox{#FFEEAA;}{\,(7-10-5)\,}{(-8)}

and when we remove the brackets in "\displaystyle {}-(-8)" we get "\displaystyle {}+8",

\displaystyle \phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=3+8
\displaystyle \phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,}-5)}{}=11.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.1:1c“