Lösung 1.1:2a

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Uttrycket består av två parenteser
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The expression consists of two brackets
<center><math>\bbox[#FFEEAA;,1.5pt]{\,(3-(7-4))\,}\cdot\bbox[#FFEEAA;,1.5pt]{\,(6-5)\,}</math></center>
<center><math>\bbox[#FFEEAA;,1.5pt]{\,(3-(7-4))\,}\cdot\bbox[#FFEEAA;,1.5pt]{\,(6-5)\,}</math></center>
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som vi kan räkna ut var för sig och sedan multiplicera ihop resultatet.
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which we can work out individually and then multiply together.
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In the expression in the first bracket, we calculate the inner bracket first
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I det första parentesuttrycket räknar vi först ut den inre parentesen
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<center><math>(3-\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}) = (3-\bbox[#FFEEAA;,1.5pt]{\,3\,})</math></center>
<center><math>(3-\bbox[#FFEEAA;,1.5pt]{\,(7-4)\,}) = (3-\bbox[#FFEEAA;,1.5pt]{\,3\,})</math></center>
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och får därför att det första parentesuttrycket blir <math>3-3=0</math>.
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and obtain that the expression in the first bracket becomes
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<math>3-3=0</math>.
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Hela uttrycket är alltså lika med 0 gånger <math>(6-5)</math>, och eftersom 0 gånger någonting alltid ger 0 så blir answer 0.
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The complete expression is 0 times <math>(6-5)</math>, and since 0 times anything is always 0 the answer is 0.
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Version vom 13:01, 13. Sep. 2008

The expression consists of two brackets

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

which we can work out individually and then multiply together. In the expression in the first bracket, we calculate the inner bracket first

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

and obtain that the expression in the first bracket becomes \displaystyle 3-3=0.

The complete expression is 0 times \displaystyle (6-5), and since 0 times anything is always 0 the answer is 0.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.1:2a“