MATHHX B
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1.2 Bogstavregning
Gymnasiematematik er mere abstrakt end folkeskolematematik. Her regner vi ofte med bogstaver, så det er vigtigt at få en god forståelse for bogstavregning fra starten af. Mange elever har svært ved bogstavregning. Det ved os
lærere godt, så spørg os og vi hjælper!
-
Eksempel 1.2.1
Her er et udtryk med bogstaver:
\[\frac {a+7}{b}-a\]
Bogstaverne \(a\) og \(b\) står for tal vi ikke kender, så man kan ikke regne videre på udtrykket.
Lad os sige, at vi får at vide at \(a=-2\) og \(b=5\). Nu kan vi regne værdien af udtrykket:
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\frac {a+7}{b}-a & =\frac {-2+7}{5} -(-2)\\[10pt] & =\frac {5}{5}+2\\[10pt] & =1+2\\ & = 3
\end{align*}
Læg mærke til parentesen om \(-2\) i første linje.
Når vi regner med bogstaver vil vi ofte undlade gangetegn. F.eks. vil vi skrive \(2a\) i stedet for \(2\cdot a\) og \(ab\) i stedet for \(a\cdot b\).
Øvelse 1.2.1
Lad \(a=3\) og \(b=-1\). Regn værdien af udtrykkene:
-
a) \(a+b\)
-
b) \(a-b\)
-
c) \(2b\)
-
d) \(ab\)
Løsning 1.2.1
-
a) \(a+b=3+(-1)=3-1=2\)
-
b) \(a-b=3-(-1)=3+1=4\)
-
c) \(2b=2\cdot (-1)=-2\)
-
d) \(ab=3\cdot (-1)=-3\)
Reduktion
Kender man ikke værdien af bogstaverne i et udtryk, kan man ikke regne det. Men nogle gange kan man forsimple det. Det kaldes at reducere.
Øvelse 1.2.2
Reducer
-
a) \(b+c-2a+a-c\)
-
b) \(a+a+a\)
-
c) \(b+x-x-b\)
Løsning 1.2.2
-
a) \(b-a\)
-
b) \(3a\)
-
c) \(0\)
Øvelse 1.2.3
Reducer:
-
a) \(ab-a+a-2b+b\)
-
b) \(bc+cb\)
-
c) \(ab+ab+ba\)
Løsning 1.2.3
-
a) \(ab-b\)
-
b) \(2bc\)
-
c) \(3ab\)
Parenteser
Vi husker at en parentes betyder, at man skal regne det, som står i parentesen først.
Når der optræder bogstaver i regnestykket, kan man ikke altid regne parentesen først. Så kan man i stedet benytte følgende regler til at hæve parentesen (dvs. fjerne parentesen):
-
• Man ganger ind i parenteser ved at gange i hvert led.
-
• Plusparenteser er overflødige og kan bare fjernes
-
• Minusparenteser hæves ved at skifte fortegn på det, der står inde i parentesen. Altså plus bliver til minus og omvendt.
Øvelse 1.2.4
Reducer
-
a) \(3(x+y)\)
-
b) \((a-b)2\)
-
c) \(2+(x+y)-3\)
-
d) \(-(v+w)\)
-
e) \(a(2b+b)-ba-(2a-2)\)
Øvelse 1.2.5
To elever diskuterer, hvordan man regner udtrykket:
\[7(3+1)\]
Elev 1 siger:
Man skal starte med at regne det inde i parentesen og derefter gange med \(7\).
Elev 2 siger:
Man skal gange parentesen ud og så reducere.
Vi slutter af med to ekstraafsnit. Husk, at man kan springe disse afsnit over, hvis man har svært ved matematik.
Ekstra: korrekt sprogbrug
I dette ekstraafsnit skal vi lære at sætte ord på de forskellige dele af et matematisk udtryk.
Sum, differens og led
En sum er det samme som et plusstykke. Det kunne være:
\[a+b+c\]
Her kaldes \(a,b,c\) for led. En sum består altså af to eller flere led adskilt af et plustegn. En differens er det samme som et minusstykke:
\[a-b\]
Her kaldes \(a\) og \(b\) også for led.
Produkt, faktorer
Et produkt er et det samme som et gangestykke. Det kunne være:
\[a\cdot b\cdot c\cdot d\]
Her kaldes \(a,b,c\) og \(d\) for faktorer. Et produkt består altså af to eller flere faktorer adskilt af gangetegn. Vi skriver ofte produkter uden gangetegn, så ovenstående produkt kunne skrives som:
\[abcd\]
Potens, grundtal og eksponent
En potens er et udtryk på formen
\[a^p\]
Her kaldes \(a\) for grundtallet og \(p\) for eksponenten.
-
Eksempel 1.2.6
Betragt udtrykket:
\[a(b+c)\]
Dette betyder \(a\cdot (b+c)\), og derfor er det et produkt bestående af faktorerne \(a\) og \((b+c)\). Den anden faktor, altså \((b+c)\), er en sum bestående af de to led \(b\) og \(c\).
Øvelse 1.2.6
Forklar, hvordan udtrykkene er opbygget.
-
a) \(2ab\)
-
b) \(a-ab\)
-
c) \(q^p\)
Løsning 1.2.6
-
a) Det er et produkt bestående af faktorerne \(2\), \(a\) og \(b\).
-
b) Det er en differens bestående af leddene \(a\) og \(ab\). Det sidste led er et produkt bestående af faktorerne \(a\) og \(b\).
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c) Det er en potens, hvor grundtallet er \(q\), og eksponenten er \(p\).
Det næste ekstraafsnit forudsætter, at du har regnet afsnit 1.1. Det er et vigtigt afsnit, hvis du gerne vil have en solid forståelse.
Ekstra: brøker med bogstaver
Når vi fremover støder på brøker, så vil de ofte være med bogstaver i stedet for tal. Det er dog præcis de samme regler, der gælder.
Forlænge og forkorte
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Eksempel 1.2.7
Vi vil forlænge brøken \(\frac {a+b}{c}\) med \(5\):
\[\frac {a+b}{c}=\frac {5(a+b)}{5c}=\frac {5a+5b}{5c}\]
Læg mærke til parentesen efter første lighedstegn. Det er en almindelig fejl, at man glemmer den. Glemmer man den, får man ikke ganget hele tælleren med \(5\), men kun første del.
Hvis der optræder flere led tæller eller nævner, skal man forkorte i hvert led.
Øvelse 1.2.7
Regn
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a) Forlæng brøken \(\frac {a+1}{b}\) med \(c\).
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b) Forkort brøken \(\frac {ab}{bc}\).
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c) Forkort brøken \(\frac {ab}{b}\)
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d) Forkort brøken \(\frac {ab+ac-a^3}{2a^2+a}\)
Plus og minus
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Eksempel 1.2.10
Vi vil regne \(\frac {2a+b}{a}+\frac {4}{b}\). Vi forlænger første brøk med \(b\) og anden brøk med \(a\):
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\frac {2a+b}{a}+\frac {4}{b}&=\frac {(2a+b)b}{ab}+\frac {4a}{ab}\\ &=\frac {(2a+b)b+4a}{ab}\\ &=\frac {2ab+b^2+4a}{ab}
\end{align*}
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Eksempel 1.2.11
Vi vil regne \(\frac {a}{b}+c\)
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\frac {a}{b}+c & = \frac {a}{b}+\frac {c}{1}\\ &=\frac {a}{b}+\frac {c\cdot b}{1\cdot b}\\ &=\frac {a}{b}+\frac {bc}{b}\\ &=\frac {a+bc}{b}
\end{align*}
Øvelse 1.2.8
Regn
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a) \(\frac {a}{b}+\frac {c}{b}\)
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b) \(\frac {a}{b}+\frac {b}{a}\)
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c) \(\frac {a}{b}-2\)
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d) \(\frac {b^2}{4a^2}-\frac {c}{a}\)
Løsning 1.2.8
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a) \(\frac {a}{b}+\frac {c}{b}=\frac {a+c}{b}\)
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b) \(\frac {a}{b}+\frac {b}{a}=\frac {a^2+b^2}{ab}\)
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c) \(\frac {a}{b}-2=\frac {a-2b}{b}\)
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d) \(\frac {b^2}{4a^2}-\frac {c}{a}=\frac {b^2-4ac}{4a^2}\)
Gange og dividere
Når man ganger brøker, som har flere led i tælleren, skal man huske at sætte parenteser.
Øvelse 1.2.9
Regn:
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a) \(\frac {a+b}{c}\cdot \frac {a}{b}\)
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b) \(\frac {a+2}{b}\cdot a\)
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c) \(\frac {\ \frac {2}{a}\ }{b}\)
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d) \(\frac {\ a\ }{\frac {2}{a}}\)
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e) \(\frac {\ \frac {3+a}{b}\ }{\frac {a}{b}}\)
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f) \(\frac {1}{\frac {1}{y}}\)
Løsning 1.2.9
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a) \(\frac {a+b}{c}\cdot \frac {a}{b}=\frac {a^2+ab}{bc}\)
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b) \(\frac {a+2}{b}\cdot a = \frac {a^2+2a}{b}\)
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c) \(\frac {\ \frac {2}{a}\ }{b}=\frac {2}{ab}\)
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d) \(\frac {\ a\ }{\frac {2}{a}}=\frac {a^2}{2}\)
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e) \(\frac {\ \frac {3+a}{b}\ }{\frac {a}{b}}= \frac {3+a}{a}\)
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f) \(\frac {1}{\frac {1}{y}}=y\)