MATHHX B
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1.5 Uligheder
En ulighed er et udsagn som indeholder et ulighedstegn.
Der findes 4 ulighedstegn:
\(<\) betyder ”mindre end”.
\(\leq \) betyder ”mindre end eller lig med”.
\(>\) betyder ”større end”.
\(\geq \) betyder ”større end eller lig med”.
Mange elever bliver ved at glemme betydningen af de forskellige ulighedstegn. Når der f.eks. står \(>\), betyder det så større eller mindre end? Er man i tvivl om, hvordan ulighedstegnet skal vende, så kan man huske, at
krokodillen er sulten, og derfor spiser det største tal:
Læg mærke til krokodillens mund som har form som ulighedstegnet \(>\). Der står altså ”7 er større end 3”. Det er en sand ulighed, da \(7\) rent faktisk er større end \(3\).
Øvelse 1.5.1
Afgør hvilke af følgende uligheder som er sande:
-
a) \(2>5\)
-
b) \(5<5\)
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c) \(11\leq 12\)
-
d) \(11\leq 11\)
Løsning 1.5.1
-
a) Falsk
-
b) Falsk
-
c) Sand
-
d) Sand
Løsning af uligheder
Man løser uligheder på samme måde som ligninger bortset fra en ting: Når man ganger eller dividerer med et negativt tal, skal man vende ulighedstegnet!
-
Eksempel 1.5.2
Vi vil løse uligheden \(2x\leq 6\). Vi dividerer med \(2\) på begge sider af ulighedstegnet, og får
\[x\leq 3\]
Altså har uligheden løsningen \(x\leq 3\).
Øvelse 1.5.2
Løs ulighederne:
-
a) \(2x\geq 6\)
-
b) \(2+x<-1\)
-
c) \(x+2>14\)
Løsning 1.5.2
-
a) \(x\geq 3\)
-
b) \(x<-3\)
-
c) \(x>12\)
-
Eksempel 1.5.3
Vi vil løse uligheden \(2x+4<6x+2(x-4)\).
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
2x+4&<6x+2(x-4) && (\text {uligheden skrevet op})\\ 2x+4&<6x+2x-8 && (\text {parentes ganget ud})\\ 2x+4&<8x-8 && (\text {højresiden reduceret})\\
2x+4-8x&<-8 && (8x \text { trukket fra på begge sider})\\ -6x+4&<-8 && (\text {venstresiden reduceret})\\ -6x&<-8-4 && (4 \text { trukket fra på
begge sider})\\ -6x&<-12 && (\text {højresiden reduceret})\\ x&>2 && (\text {delt med } -6 \text { på begge sider})\\
\end{align*}
Læg mærke til, hvordan vi har vendt ulighedstegnet til sidst, hvor vi dividerer med \(-6\).
Øvelse 1.5.3
Løs ulighederne og læs facit op (læs inde i dit hoved, så du ikke forstyrre hele klassen).
Løsning 1.5.3
-
a) \(x\leq -7\). Læses ”\(x\) er mindre end eller lig med \(-7\)”.
-
b) \(x>-6\). Læses ”\(x\) er større end \(-6\)”.
-
c) \(x\geq -5{,}5\). Læses ”\(x\) er større end eller lig med \(-5{,}5\)”.
-
d) \(x<3\). Læses ”\(x\) er mindre end \(3\)”.
-
e) \(x\leq -2{,}25\). Læses ”\(x\) er mindre end eller lig med \(-2{,}25\)”.
-
f) \(x<-0{,}75\). Læses ”\(x\) er mindre end \(-0{,}75\)”.
Øvelse 1.5.4
Uligheder med brøker løses på tilsvarende måde, som man løser ligninger med brøker. Hvis du ikke har læst ekstraafsnittet med brøker, så spring denne øvelse over.
Løs ulighederne:
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a) \(3\geq \frac {x}{-5}\)
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b) \(\frac {6}{x}<2\). Forudsæt at \(x\) er positiv.
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c) \(\frac {6}{x}<2\). Forudsæt at \(x\) er negativ (svær)
Ekstra: Løsningsmængde og grundmængde
Ligesom ligninger har løsningsmængde og grundmængde har uligheder det også. Det fungere helt tilsvarende.
-
Eksempel 1.5.4
Vi bestemmer løsningsmængden for uligheden \(2x+1<3\). Det ses nemt, at løsningen er
\[x<1\]
De \(x\)-værdier, som opfylder denne ulighed, vil ligge i intervallet:
\[]-\infty ;1[\]
Løsningsmængden \(L\) er dermed:
\[L=]-\infty ;1[\]
Øvelse 1.5.5
Betragt uligheden \(4x\geq 2x+10\).
Løsning 1.5.5
-
a) \(L=[5;\infty [\)
-
b) \(G=\mathbb {R}\)