Breuken met letters

Leerdoelen

•Je kunt uitleggen wat een breuk met letters is.

•Je kunt breuken met letters vereenvoudigen door teller en noemer te herleiden.

•Je kunt breuken met letters optellen en aftrekken door de noemers gelijknamig te maken.

•Je kunt breuken met letters vermenigvuldigen en delen volgens de juiste regels.

•Je kunt herkennen wanneer een breuk verder kan worden vereenvoudigd.

Breuken met letters vereenvoudigen

Neem de breuk\frac{12xy}{2x}\frac{12xy}{2x}\frac{12xy}{2}\frac{12xy}{}\frac{12xy}[\frac{12xy}[2\frac{12xy}[2x\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}}\large{\frac{12xy}{2x}} \large{\frac{12xy}{2x}} . Dit is ook te zien als\frac{12\cdot x\cdot y}{2 \cdot x}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\frac{12\cdot x\cdot y}{2 \cdot x}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}}\large{\frac{12 \cdot x \cdot y}{2 \cdot x}} \large{\frac{12 \cdot x \cdot y}{2 \cdot x}} .

Deze breuk is nu te herleiden door de teller en de noemer systematisch door dezelfde getallen en letters te delen.

1.Deel eerst de 12 door 2, wat 6 oplevert.

2.Vervolgens deel je x door x, wat 1 oplevert en uiteindelijk deel je i door 1, wat i oplevert.

3.Op deze manier hou je de breuk 6 \cdot 1 \cdot y over, oftewel 6 \cdot y .

Probeer dit nu ook eens met de breuk\frac{8a}{2ab}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}}\large{\frac{8a}{2ab}} \large{\frac{8a}{2ab}} .

1.Begin door 8 te delen door 2, wat 4 oplevert

2.Vervolgens a door a te delen, wat 1 oplevert

3.Tenslotte 1 door b te delen. Dat geeft de breuk\frac{4}{b}\large{\frac{4}{b}}\frac{4}{b}\large{\frac{4}{b}}\large{\frac{4}{b}}\large{\frac{4}{b}}\large{\frac{4}{b}}\large{\frac{4}{b}}\large{\frac{4}{b}}\large{\frac{4}{b}}\large{\frac{4}{b}}\large{\frac{4}{b}}\large{\frac{4}{b}}\large{\frac{4}{b}} \large{\frac{4}{b}} .

Breuken met letters optellen en aftrekken

In deze breuk \frac{2}{5x}+\frac{3}{5x}\large{\frac{2}{5x}}\frac{2}{5x}+\frac{3}{5x}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}}\large{\frac{2}{5x}} \large{\frac{2}{5x}} ⁣, zijn de noemers gelijk. De tellers moeten bij elkaar opgeteld worden en de noemer blijft hetzelfde. De som hiervan is dan\frac{5}{5x}\large{\frac{5}{5x}}\frac{5}{5x}\large{\frac{5}{5x}}\large{\frac{5}{5x}}\large{\frac{5}{5x}}\large{\frac{5}{5x}}\large{\frac{5}{5x}}\large{\frac{5}{5x}}\large{\frac{5}{5x}}\large{\frac{5}{5x}}\large{\frac{5}{5x}}\large{\frac{5}{5x}}\large{\frac{5}{5x}}\large{\frac{5}{5x}} \large{\frac{5}{5x}} . Deze breuk kunnen we ook nog vereenvoudigen tot\frac{1}{x}\large{\frac{1}{x}}\frac{1}{x}\large{\frac{1}{x}}\large{\frac{1}{x}}\large{\frac{1}{x}}\large{\frac{1}{x}}\large{\frac{1}{x}}\large{\frac{1}{x}}\large{\frac{1}{x}}\large{\frac{1}{x}}\large{\frac{1}{x}}\large{\frac{1}{x}}\large{\frac{1}{x}} \large{\frac{1}{x}} .

Probeer dit principe ook eens met de breuk\frac{2ab}{3ac}+\frac{4a}{3c}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}} \large{\frac{2ab}{3ac}} .

1.Noemers zijn niet hetzelfde dus deze eerst gelijk maken door beide delen van de tweede fractie met een 'a' te vermenigvuldigen. Dit geeft de breuk\frac{2ab}{3ac}+\frac{4a^{2}}{3ac}\large{\frac{}{3ac}}\frac{2ab}{3ac}+\frac{4a^{2}}{3ac}\large{\frac{2ab}{3ac}}\frac{2ab}{3ac}+\frac{4a^{2}}{3ac}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}}\large{\frac{2ab}{3ac}} \large{\frac{2ab}{3ac}}

2. Nu de noemers hetzelfde zijn, kunnen we de tellers optellen en dit vereenvoudigen tot\frac{2b+4a}{3c}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}}\large{\frac{2b + 4a}{3c}} \large{\frac{2b + 4a}{3c}} .

Breuken met letters vermenigvuldigen en delen

Bij het vermenigvuldigen van breuken met letters volg je een eenvoudige regel: je vermenigvuldigt de teller van de eerste breuk met de teller van de tweede breuk en de noemer van de eerste breuk met de noemer van de tweede breuk.

\frac{a}{b}\cdot\frac{3b}{5a}\frac{a}{b}\cdot\frac{3b}{5}\frac{a}{b}\cdot\frac{3b}{\placeholder{}}\frac{a}{b}\cdot3b\frac{a}{b}3b\frac{a}{b}53b\frac{a}{b}5a3b\frac{a}{\placeholder{}}5a3ba5a3b

Stap 1: teller × teller en noemer × noemer

{\frac{a}{b}\cdot\frac{3b}{5a}}=\frac{3ab}{5ab}{\frac{a}{\placeholder{}}\cdot\frac{3b}{5a}}=\frac{3ab}{5ab}{a\cdot\frac{3b}{5a}}=\frac{3ab}{5ab}{a\cdot\frac{3b}{\cdot5a}}=\frac{3ab}{5ab}{a\cdot\frac{3b}{b\cdot5a}}=\frac{3ab}{5ab}{a\cdot\frac{3b}{\placeholder{}}}=\frac{3ab}{5ab}{a\cdot\frac{3b}{\placeholder{}}}{b \cdot5a}=\frac{3ab}{5ab}{a\cdot3b}{b \cdot5a}=\frac{3ab}{5ab}{a\cdot\frac{3b}{\placeholder{}}}{b \cdot5a}=\frac{3ab}{5ab}

Stap 2: vereenvoudigen

\frac{3ab}{5ab}=\frac35\frac{3ab}{5ab}=\frac{35}{5}\frac{3ab}{5ab}=\frac{35}{\placeholder{}}\frac{3ab}{5ab}=35\frac{3ab}{5ab}b=35\frac{3ab}{5ab}ab=35\frac{3ab}{5ab}5ab=35\frac{3ab}{5a}5ab=35\frac{3ab}{5}5ab=35\frac{3ab}{\placeholder{}}5ab=35

Breuken delen

Delen door een breuk betekent dat je vermenigvuldigt met het omgekeerde van die breuk. Dit betekent dat je de teller en noemer van de tweede breuk verwisselt en vervolgens vermenigvuldigt zoals bij gewone vermenigvuldiging.

Voorbeeld

Neem de breuken:

\frac{3a}{4b}\div\frac{5c}{6d}\frac{3a}{4b}\div\frac{5c}{6d}d\frac{3a}{4b}\div\frac{5c}{6d}6d\frac{3a}{4b}\div\frac{5c}{6}6d\frac{3a}{4b}\div\frac{5c}{\placeholder{}}6d\frac{3a}{4b}\div5c6d\frac{3a}{4b}\div5c.6d\frac{3a}{4b}\div5c6d\frac{3a}{4b}b\div5c6d\frac{3a}{4b}4b\div5c6d\frac{3a}{4}4b\div5c6d\frac{3a}{\placeholder{}}4b\div5c6d3a4b\div5c6d3a4b\div5c63a4b\div5c6d

Stap 1: schrijf als vermenigvuldiging met het omgekeerde:

\frac{3a}{4b}\times\frac{6d}{5c}\frac{3a}{4b}\times\frac{6d}{5c}c\frac{3a}{4b}\times\frac{6d}{5c}5c\frac{3a}{4b}\times\frac{6d}{5}5c\frac{3a}{4b}\times\frac{6d}{\placeholder{}}5c\frac{3a}{4b}\times6d5c\frac{3a}{4b}b\times6d5c\frac{3a}{4b}4b\times6d5c\frac{3a}{4}4b\times6d5c\frac{3a}{\placeholder{}}4b\times6d5c

Stap 2: teller × teller en noemer × noemer

\frac{3a\cdot6d}{4b\cdot5c}=\frac{18ad}{20bc}\frac{3a\cdot6d}{4b\cdot5c}=\frac{18ad}{20b}\frac{3a\cdot6d}{4b\cdot5c}=\frac{18ad}{20}\frac{3a\cdot6d}{4b\cdot5c}=\frac{18ad}{2}\frac{3a\cdot6d}{4b\cdot5c}=\frac{18ad}{\placeholder{}}\frac{3a\cdot6d}{4b\cdot5c}=18ad\frac{3a\cdot6d}{4b\cdot5c}=18a\frac{3a\cdot6d}{4b\cdot5c}=18\frac{3a\cdot6d}{4b\cdot5c}=1\frac{3a\cdot6d}{4b\cdot5c}=\frac{3a\cdot6d}{4b\cdot5c}\frac{3a\cdot6d}{4b\cdot5}\frac{3a\cdot6d}{4b\cdot}\frac{3a\cdot6d}{4b}\frac{3a\cdot6d}{4}\frac{3a\cdot6d}{\placeholder{}}\frac{3a\cdot6}{\placeholder{}}\frac{3a\cdot}{\placeholder{}}\frac{3a}{\placeholder{}}3a3

Stap 3: vereenvoudigen

\frac{18ad}{20bc}=\frac{9ad}{10bc}\frac{18ad}{20bc}=\frac{9ad}{10b}\frac{18ad}{20bc}=\frac{9ad}{10}\frac{18ad}{20bc}=\frac{9ad}{1}\frac{18ad}{20bc}=\frac{9ad}{}\frac{18ad}{20bc}=\frac{9ad}{`}\frac{18ad}{20bc}=\frac{9ad}{``}\frac{18ad}{20bc}=\frac{9ad}{``1}\frac{18ad}{20bc}=\frac{9ad}{``10}\frac{18ad}{20bc}=\frac{9ad}{``1}\frac{18ad}{20bc}=\frac{9ad}{``}\frac{18ad}{20bc}=\frac{9ad}{``}\frac{18ad}{20bc}=\frac{9ad}{\placeholder{}}\frac{18ad}{20bc}=9ad\frac{18ad}{20bc}=9a\frac{18ad}{20bc}=9\frac{18ad}{20bc}=\frac{18ad}{20bc}\frac{18ad}{20b}\frac{18ad}{20}\frac{18ad}{2}\frac{18ad}{\placeholder{}}18ad18a181

Uitleg:

Draai de tweede breuk om.

Vermenigvuldig teller met teller en noemer met noemer.

Vereenvoudig factoren die in teller en noemer voorkomen.