MATHHX B

MATHHX B

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1.5 Uligheder

En ulighed er ligesom en ligning bortset fra at den indeholder et ulighedstegn i stedet for et lighedstegn

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 glemmer betydningen af hvordan ulighedstegnet vender.. Når der står \(>\), betyder det så større eller mindre end? Er man i tvivl om hvordan de skal vende, så kan man huske at krokodillen altid spiser det største tal:

(image)

Læg mærke til krokodillens mund som har form som tegnet \(>\). 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) \(5<5\)

  • c) \(11\leq 12\)

  • d) \(11\leq 11\)

Løsning 1.5.1

  • a) Falsk

  • b) Falsk

  • c) Sand

  • d) Sand

Løsning af uligheder

At løse en ulighed er som at løse en ligning bortset fra en ting: Når man ganger eller dividerer med et negativt tal, skal man vende ulighedstegnet!

  • Eksempel 1.5.1
    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 1.5.2
    Vi vil løse uligheden \(2x+4<6x+2(x-4)\).

    \begin{align*} 2x+4&<6x+2(x-4) && (\text {Skriver uligheden op})\\ 2x+4&<6x+2x-8 && (\text {Ganger parentes ud})\\ 2x+4&<8x-8 && (\text {Reducerer højresiden})\\ 2x+4-8x&<-8 && \text {(Trækker } 8x \text { fra på begge sider)}\\ -6x+4&<-8 && (\text {Reducerer venstresiden})\\ -6x&<-8-4 && \text {(Trækker } 4 \text { fra på begge sider)}\\ -6x&<-12 && (\text {Reducerer højresiden})\\ x&>2 && \text {(Deler med } -6 \text { fra 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:

  • a) \(-x\geq 7\)

  • b) \(2-x<8\)

  • c) \((x-2)\cdot 3\leq 5(x+1)\)

  • d) \(2(2+x)-(x-1)<8\)

  • e) \(0\geq 5x+10-(x+1)\)

  • f) \(-(x+3)\cdot 2>(5+1)\cdot x\)

Løsning 1.5.3

  • a) \(x\leq -7\)

  • b) \(x>-6\)

  • c) \(x\geq -5{,}5\)

  • d) \(x<3\)

  • e) \(x\leq -2{,}25\)

  • f) \(x<-0{,}75\)

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:

  • a) \(3\geq \frac {x}{-5}\)

  • b) \(\frac {6}{x}<2\). Forudsæt at \(x\) er positiv.

  • c) \(\frac {6}{x}<2\). Forudsæt at \(x\) er negativ (svær)

Løsning 1.5.4

  • a) \(x\geq -15\)

  • b) \(x>3\)

  • c) \(x<0\)

Dobbeltuligheder (A)

En dobbeltulighed er en ulighed med to ulighedstegn. Den kunne f.eks. se således ud:

\[3<2x+1<9\]

Tænker man lidt over det, så er det klart, at sådan en dobbeltulighed bare er en kompakt måde at skrive to uligheder, nemlig:

\[3<2x+1\qquad \text {og}\qquad 2x+1<9\]

De to uligheder kan vi løse på sædvanlig vis (gør det!), og det giver:

\[x>1\qquad \text {og}\qquad x<4\]

Så \(x\) skal altså være større end \(1\), men mindre end \(4\). Dvs. \(x\) skal ligge imellem \(1\) og \(4\). Det kan vi også skrive som:

\[1<x<4\]

og dette er så løsningen til dobbeltuligheden.

Øvelse 1.5.5

Løs ulighederne. De to sidste er måske lidt svære, hvis du ikke har regnet øvelsen om uligheder med brøker.

  • a) \(1<x-1<5\)

  • b) \(x+1\leq 2x<x+2\)

  • c) \(-1{,}5\leq -\frac {a}{400}\leq -0{,}25\)

  • d) \(-1{,}5\leq -\frac {300}{b}\leq -0{,}25\) (forudsæt at \(b>0\)).

Løsning 1.5.5

  • a) \(2<x<6\)

  • b) \(1\leq x<2\)

  • c) \(100\leq a\leq 600\)

  • d) \(200\leq b \leq 1200\)

Ekstra: Løsningsmængde og grundmængde

Ligesom ligninger har løsningsmængde og grundmængde har uligheder det også. Det fungere helt tilsvarende.

  • Eksempel 1.5.3
    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 1.5.6

Opskriv løsningsmængden for ulighederne i øvelse 1.5.5.

  • a) \(1<x-1<5\)

  • b) \(x+1\leq 2x<x+2\)

  • c) \(-1{,}5\leq -\frac {a}{400}\leq -0{,}25\)

  • d) \(-1{,}5\leq -\frac {300}{b}\leq -0{,}25\) (forudsæt at \(b>0\)).

Løsning 1.5.6

  • a) \(]2;6[\)

  • b) \([1;2[\)

  • c) \([100;600]\)

  • d) \([200;1200]\)