2.1 Algebraische Ausdrücke

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Regerate images and tabs)
Zeile 2: Zeile 2:
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
| style="border-bottom:1px solid #797979" width="5px" |  
| style="border-bottom:1px solid #797979" width="5px" |  
-
{{Vald flik|[[2.1 Algebraiska uttryck|Teori]]}}
+
{{Vald flik|[[2.1algebraic expressions|Teori]]}}
-
{{Ej vald flik|[[2.1 Övningar|Övningar]]}}
+
{{Ej vald flik|[[2.1 Övningar|Exercises]]}}
| style="border-bottom:1px solid #797979" width="100%"|  
| style="border-bottom:1px solid #797979" width="100%"|  
|}
|}
{{Info|
{{Info|
-
'''Innehåll:'''
+
'''Content:'''
-
*Distributiva lagen
+
* Distributive law
-
*Kvadreringsreglerna
+
* Squaring rules
-
*Konjugatregeln
+
*Difference of two squares
-
*Rationella uttryck
+
* Rational expression
}}
}}
{{Info|
{{Info|
-
'''Lärandemål:'''
+
'''Learning outcomes:'''
-
Efter detta avsnitt ska du ha lärt dig att:
+
After this section, you will have learned how to:
-
*Förenkla komplicerade algebraiska uttryck.
+
*Simplify complex algebraic expression.
-
*Faktorisera uttryck med kvadreringsreglerna och konjugatregeln.
+
*Factorise expressions using squaring rules and and the difference of two squares rule.
-
*Utveckla uttryck med kvadreringsreglerna och konjugatregeln.
+
*Expand expressions using squaring rules and and the difference of two squares rule.
}}
}}
-
== Distributiva lagen ==
+
== Distributive Law ==
[[Bild:miniräknare_skämt.gif|right]]
[[Bild:miniräknare_skämt.gif|right]]
-
Den distributiva lagen anger hur man multiplicerar in en faktor i en parentes.
+
The distributive law specifies how to multiply a bracketed expression by a factor.
<center>{{:2.1 - Figur - Distributiva lagen}}</center>
<center>{{:2.1 - Figur - Distributiva lagen}}</center>
<div class="exempel">
<div class="exempel">
-
'''Exempel 1'''
+
''' Example 1'''
<ol type="a">
<ol type="a">
Zeile 47: Zeile 47:
</div>
</div>
-
Med den distributiva lagen kan vi också förstå hur vi kan hantera minustecken framför parentesuttryck. Regeln säger att en parentes med ett minustecken framför kan tas bort om alla termer inuti parentesen byter tecken.
+
Using the distributive law, we can also see how to tackle
 +
a minus sign in front of a bracketed expression.
 +
The rule says that bracket with a minus sign in front can be
 +
eliminated if all the terms inside the brackets, switch signs.
<div class="exempel">
<div class="exempel">
-
'''Exempel 2'''
+
''' Example 2'''
<ol type="a">
<ol type="a">
Zeile 56: Zeile 59:
<li><math>-(x^2-x) = (-1) \cdot (x^2-x) = (-1)x^2 -(-1)x
<li><math>-(x^2-x) = (-1) \cdot (x^2-x) = (-1)x^2 -(-1)x
= -x^2 +x</math><br/>
= -x^2 +x</math><br/>
-
:där vi i sista ledet använt att <math>-(-1)x = (-1)(-1)x = 1\cdot x = x\,\mbox{.}</math></li>
+
where we have in the final step used <math>-(-1)x = (-1)(-1)x = 1\cdot x = x\,\mbox{.}</math></li>
<li><math>-(x+y-y^3) = (-1)\cdot (x+y-y^3) = (-1)\cdot x
<li><math>-(x+y-y^3) = (-1)\cdot (x+y-y^3) = (-1)\cdot x
+ (-1) \cdot y -(-1)\cdot y^3</math><br/>
+ (-1) \cdot y -(-1)\cdot y^3</math><br/>
Zeile 65: Zeile 68:
</div>
</div>
-
Om den distributiva lagen används baklänges så sägs vi faktorisera uttrycket. Ofta försöker man bryta ut en så stor faktor som möjligt.
+
If the distributive law is applied in backwards we say we factor the expression.
 +
One often would like to to factorise out as large a numerical factor as possible.
<div class="exempel">
<div class="exempel">
-
'''Exempel 3'''
+
''' Example 3'''
<ol type="a">
<ol type="a">
Zeile 79: Zeile 83:
-
== Kvadreringsreglerna ==
+
== Squaring rules ==
-
Den distributiva lagen behöver ibland användas upprepade gånger för att behandla större uttryck. Om vi betraktar
+
The distributive law occasionally has to be used repeatedly to deal with larger expression.
 +
If we consider
{{Fristående formel||<math>(a+b)(c+d)</math>}}
{{Fristående formel||<math>(a+b)(c+d)</math>}}
-
och ser <math>a+b</math> som en faktor som multipliceras in i parentesen (c+d) så får vi
+
and regard <math>a+b</math> as a factor that multiplies the bracketed expression(c+d) we get
{{Fristående formel||<math>\eqalign{
{{Fristående formel||<math>\eqalign{
Zeile 94: Zeile 99:
&= (a+b)\,c + (a+b)\,d\mbox{.}}</math>}}
&= (a+b)\,c + (a+b)\,d\mbox{.}}</math>}}
-
Sedan kan <math>c</math> och <math>d</math> multipliceras in i respektive parentes
+
Then the <math>c</math> and the <math>d</math>are multiplied into their respective brackets,
{{Fristående formel||<math>(a+b)c + (a+b)d = ac + bc + ad + bd \, \mbox{.}</math>}}
{{Fristående formel||<math>(a+b)c + (a+b)d = ac + bc + ad + bd \, \mbox{.}</math>}}
-
Ett minnesvärt sätt att sammanfatta formeln är:
+
A mnemonic for this formula is:
<center>{{:2.1 - Figur - Distributiva lagen två gånger}}</center>
<center>{{:2.1 - Figur - Distributiva lagen två gånger}}</center>
<div class="exempel">
<div class="exempel">
-
'''Exempel 4'''
+
''' Example 4'''
<ol type="a">
<ol type="a">
Zeile 115: Zeile 120:
= 2-x-2x+x^2</math><br/>
= 2-x-2x+x^2</math><br/>
<math>\phantom{(1-x)(2-x)}{}=2-3x+x^2</math>
<math>\phantom{(1-x)(2-x)}{}=2-3x+x^2</math>
-
:där vi använt att <math>-x\cdot (-x) = (-1)x \cdot (-1)x = (-1)^2 x^2 = 1\cdot x^2 = x^2</math>.
+
   where we have used <math>-x\cdot (-x) = (-1)x \cdot (-1)x = (-1)^2 x^2 = 1\cdot x^2 = x^2</math>.
</ol>
</ol>
</div>
</div>
-
Två viktiga specialfall av ovanstående formel är när <math>a+b</math> och <math>c+d</math> är samma uttryck
+
Two important special cases of the above formula is when <math>a+b</math> and <math>c+d</math> are the same expression
<div class="regel">
<div class="regel">
-
'''Kvadreringsreglerna'''
+
'''Squaring rules '''
{{Fristående formel||<math>(a+b)^2 = a^2 +2ab + b^2</math>}}
{{Fristående formel||<math>(a+b)^2 = a^2 +2ab + b^2</math>}}
{{Fristående formel||<math>(a-b)^2 = a^2 -2ab + b^2</math>}}
{{Fristående formel||<math>(a-b)^2 = a^2 -2ab + b^2</math>}}
</div>
</div>
-
Dessa formler kallas för första och andra kvadreringsregeln.
+
These formulas are called the first and second squaring rules
<div class="exempel">
<div class="exempel">
-
'''Exempel 5'''
+
''' Example 5'''
<ol type="a">
<ol type="a">
<li><math>(x+2)^2 = x^2 + 2\cdot 2x+ 2^2 = x^2 +4x +4</math></li>
<li><math>(x+2)^2 = x^2 + 2\cdot 2x+ 2^2 = x^2 +4x +4</math></li>
<li><math>(-x+3)^2 = (-x)^2 + 2\cdot 3(-x) + 3^2 = x^2 -6x +9</math> <br>
<li><math>(-x+3)^2 = (-x)^2 + 2\cdot 3(-x) + 3^2 = x^2 -6x +9</math> <br>
-
:där <math>(-x)^2 = ((-1)x)^2 = (-1)^2 x^2 = 1 \cdot x^2 = x^2\,\mbox{.}</math></li>
+
: where <math>(-x)^2 = ((-1)x)^2 = (-1)^2 x^2 = 1 \cdot x^2 = x^2\,\mbox{.}</math></li>
<li><math>(x^2 -4)^2 = (x^2)^2 - 2 \cdot 4x^2 + 4^2
<li><math>(x^2 -4)^2 = (x^2)^2 - 2 \cdot 4x^2 + 4^2
= x^4 -8x^2 +16</math></li>
= x^4 -8x^2 +16</math></li>
Zeile 151: Zeile 156:
</div>
</div>
-
Kvadreringsreglerna används också i omvänd riktning för att faktorisera uttryck.
+
The squaring rules are also used in the reverse direction to factorise expressions.
<div class="exempel">
<div class="exempel">
-
'''Exempel 6'''
+
''' Example 6'''
<ol type="a">
<ol type="a">
Zeile 165: Zeile 170:
-
== Konjugatregeln ==
+
== Difference of two squares ==
-
Ett tredje specialfall av den första formeln i förra avsnittet är konjugatregeln
+
A third special case of the first formula in the last section is the difference of two squares rule.
<div class="regel">
<div class="regel">
Zeile 174: Zeile 179:
</div>
</div>
-
Denna formel kan vi få fram direkt genom att utveckla vänsterledet
+
This formula can be obtained directly by expanding the left hand side
{{Fristående formel||<math>(a+b)(a-b)
{{Fristående formel||<math>(a+b)(a-b)
Zeile 182: Zeile 187:
<div class="exempel">
<div class="exempel">
-
'''Exempel 7'''
+
''' Example 7'''
<ol type="a">
<ol type="a">
Zeile 195: Zeile 200:
-
== Rationella uttryck ==
+
== Rational expressions==
-
Räkning med algebraiska uttryck som innehåller bråk liknar till stor del vanlig bråkräkning.
+
Calculations of fractions containing algebraic expressions are largely similar to ordinary calculations with fractions.
-
Multiplikation och division av bråkuttryck följer samma räkneregler som gäller för vanliga bråktal,
+
Multiplication and division of fractions containing algebraic expressions follow the same rules that apply to ordinary fractions,
<div class="regel">
<div class="regel">
Zeile 210: Zeile 215:
<div class="exempel">
<div class="exempel">
-
'''Exempel 8'''
+
''' Example 8'''
<ol type="a">
<ol type="a">
Zeile 223: Zeile 228:
</div>
</div>
-
Förlängning av ett bråkuttryck innebär att vi multiplicerar täljare och nämnare med samma faktor
+
A fractional expression can have its numerator and denominator multiplied by the same factor
{{Fristående formel||<math>\frac{x+2}{x+1}
{{Fristående formel||<math>\frac{x+2}{x+1}
Zeile 230: Zeile 235:
= \dots</math>}}
= \dots</math>}}
-
Förkortning av ett bråkuttryck innebär att vi stryker faktorer som täljaren och nämnaren har gemensamt
+
The opposite of this, is cancellation, where we delete factors that the numerator and denominator have in common
-
 
+
{{Fristående formel||<math>\frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4) }
{{Fristående formel||<math>\frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4) }
= \frac{(x+2)(x+4)}{(x+1)(x+4)}
= \frac{(x+2)(x+4)}{(x+1)(x+4)}
Zeile 237: Zeile 241:
<div class="exempel">
<div class="exempel">
-
'''Exempel 9'''
+
''' Example 9'''
<ol type="a">
<ol type="a">
Zeile 250: Zeile 254:
</div>
</div>
-
När bråkuttryck adderas eller subtraheras behöver de, om så är nödvändigt, förlängas så att de får samma nämnare innan täljarna kan kombineras ihop,
+
When fractional expressions are added or subtracted, they may need to be converted so that they have the same denominator before the numerators can be combined together,
 +
 
{{Fristående formel||<math>\frac{1}{x} - \frac{1}{x-1}
{{Fristående formel||<math>\frac{1}{x} - \frac{1}{x-1}
Zeile 258: Zeile 263:
= \frac{-1}{x(x-1)} \; \mbox{.}</math>}}
= \frac{-1}{x(x-1)} \; \mbox{.}</math>}}
-
Ofta försöker man förlänga med så lite som möjligt för att underlätta räknandet. Minsta gemensamma nämnare (MGN) är den gemensamma nämnare som innehåller minst antal faktorer.
+
One normally tries to convert the fractions by multiplying the numerators and denominators by minimal factors to facilitate the calculations. The lowest common denominator (LCD) is the common denominator which contains the least number of factors.
<div class="exempel">
<div class="exempel">
-
'''Exempel 10'''
+
''' Example 10'''
<ol type="a">
<ol type="a">
-
<li><math>\frac{1}{x+1} + \frac{1}{x+2}\quad</math> har <math>\ \text{MGN}
+
<li><math>\frac{1}{x+1} + \frac{1}{x+2}\quad</math> has <math>\ \text{LCD}
= (x+1)(x+2)</math> <br><br>
= (x+1)(x+2)</math> <br><br>
-
Förläng den första termen med <math>(x+2)</math> och den andra termen med <math>(x+1)</math>
+
Convert the first term using <math>(x+2)</math> and the second term using <math>(x+1)</math>
{{Fristående formel||<math>\begin{align*}
{{Fristående formel||<math>\begin{align*}
\frac{1}{x+1} + \frac{1}{x+2}
\frac{1}{x+1} + \frac{1}{x+2}
Zeile 273: Zeile 278:
= \frac{2x+3}{(x+1)(x+2)}\:\mbox{.}
= \frac{2x+3}{(x+1)(x+2)}\:\mbox{.}
\end{align*}</math>}}</li>
\end{align*}</math>}}</li>
-
<li><math>\frac{1}{x} + \frac{1}{x^2}\quad</math> har <math>\ \text{MGN}
+
<li><math>\frac{1}{x} + \frac{1}{x^2}\quad</math> has <math>\ \text{LCD}
= x^2</math><br><br>
= x^2</math><br><br>
-
Vi behöver bara förlänga den första termen för att få en gemensam nämnare
+
We only need to convert the first term to get a common denominator
{{Fristående formel||<math>\frac{1}{x} + \frac{1}{x^2}
{{Fristående formel||<math>\frac{1}{x} + \frac{1}{x^2}
= \frac{x}{x^2} + \frac{1}{x^2}
= \frac{x}{x^2} + \frac{1}{x^2}
= \frac{x+1}{x^2}\,\mbox{.}</math>}}</li>
= \frac{x+1}{x^2}\,\mbox{.}</math>}}</li>
-
<li><math>\frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}\quad</math> har <math>\
+
<li><math>\frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}\quad</math> has <math>\
-
\text{MGN}= x^2(x+1)^2(x+2)</math><br><br>
+
\text{LCD}= x^2(x+1)^2(x+2)</math><br><br>
-
Den första termen förlängs med <math>x(x+2)</math> medan den andra termen förlängs med <math>(x+1)^2</math>
+
The first term is converted using <math>x(x+2)</math> while the other term is converted using <math>(x+1)^2</math>
{{Fristående formel||<math>\begin{align*}
{{Fristående formel||<math>\begin{align*}
\frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}
\frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}
Zeile 291: Zeile 296:
&= \frac{-1}{x^2(x+1)^2(x+2)}\,\mbox{.}
&= \frac{-1}{x^2(x+1)^2(x+2)}\,\mbox{.}
\end{align*}</math>}}</li>
\end{align*}</math>}}</li>
-
<li><math>\frac{x}{x+1} - \frac{1}{x(x-1)} -1 \quad</math> har <math>\
+
<li><math>\frac{x}{x+1} - \frac{1}{x(x-1)} -1 \quad</math> has <math>\
-
\text{MGN}=x(x-1)(x+1)</math><br><br>
+
\text{LCD}=x(x-1)(x+1)</math><br><br>
-
Vi förlänger alla termer så att de får den gemensamma nämnaren <math>x(x-1)(x+1)</math>
+
  We must convert all the terms so that they have the common denominator <math>x(x-1)(x+1)</math>
{{Fristående formel||<math>\begin{align*}
{{Fristående formel||<math>\begin{align*}
\frac{x}{x+1} - \frac{1}{x(x-1)} -1
\frac{x}{x+1} - \frac{1}{x(x-1)} -1
Zeile 307: Zeile 312:
</div>
</div>
-
Vid förenkling av större uttryck är det ofta nödvändigt att både förlänga och förkorta i steg. Eftersom förkortning förutsätter att vi kan faktorisera uttryck är det viktigt att försöka behålla uttryck (t.ex. nämnare) faktoriserade och inte utveckla något som vi senare behöver faktorisera.
+
To simplify large expressions, it is often necessary to both cancel factors and multiply numerators and denominators by factors. As cancellation implies that we have performed factorisations, it is obvious we should try to keep expressions (such as the denominator) factorised and not expand something that we will later need to factorise.
 +
 
<div class="exempel">
<div class="exempel">
-
'''Exempel 11'''
+
''' Example 11'''
<ol type="a">
<ol type="a">
Zeile 340: Zeile 346:
-
[[2.1 Övningar|Övningar]]
+
[[2.1 Övningar|Exercises]]
<div class="inforuta" style="width:580px;">
<div class="inforuta" style="width:580px;">
-
'''Råd för inläsning'''
+
'''Study advice'''
-
'''Grund- och slutprov'''
+
'''The basic and final tests''
-
Efter att du har läst texten och arbetat med övningarna ska du göra grund- och slutprovet för att bli godkänd på detta avsnitt. Du hittar länken till proven i din student lounge.
+
After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
-
'''Tänk på att:'''
+
'''Keep in mind that: '''
-
Var noggrann. Om du gör ett fel på ett ställe så kommer resten av uträkningen också vara fel.
+
Be careful. If you make a mistake somewhere the rest of the calculation will be wrong.
-
Använd många mellanled. Om du är osäker på en uträkning utför då hellre enkla steg än ett stort steg.
+
Use many intermediate steps . If you are unsure of a calculation do it in many small steps rather than one big step.
-
Utveckla inte i onödan. Du kan vid ett senare tillfälle vara tvungen att faktorisera tillbaka.
+
Do not expand unnecessarily. You later may be forced to faktorise what you earlier expanded.
-
'''Lästips'''
+
'''Reviews
 +
'''
-
[http://en.wikipedia.org/wiki/Algebra Läs mer om algebra på engelska Wikipedia]
+
[http://en.wikipedia.org/wiki/Algebra Learn more about algebra in the English Wikipedia ]
-
[http://www.jamesbrennan.org/algebra/ Understanding Algebra - engelsk textbok på nätet]
+
[http://www.jamesbrennan.org/algebra/ Understanding Algebra - English text on the Web ]

Version vom 12:31, 10. Jul. 2008

 

Vorlage:Vald flik Vorlage:Ej vald flik

 

Content:

  • Distributive law
  • Squaring rules
  • Difference of two squares
  • Rational expression

Learning outcomes:

After this section, you will have learned how to:

  • Simplify complex algebraic expression.
  • Factorise expressions using squaring rules and and the difference of two squares rule.
  • Expand expressions using squaring rules and and the difference of two squares rule.

Distributive Law

The distributive law specifies how to multiply a bracketed expression by a factor.

2.1 - Figur - Distributiva lagen

Example 1

  1. \displaystyle 4(x+y) = 4x + 4y
  2. \displaystyle 2(a-b) = 2a -2b
  3. \displaystyle x \left(\frac{1}{x} + \frac{1}{x^2} \right) = x\cdot \frac{1}{x} + x \cdot \frac{1}{x^2} = \frac{\not{x}}{\not{x}} + \frac{\not{x}}{x^{\not{2}}} = 1 + \frac{1}{x}
  4. \displaystyle a(x+y+z) = ax + ay + az

Using the distributive law, we can also see how to tackle a minus sign in front of a bracketed expression. The rule says that bracket with a minus sign in front can be eliminated if all the terms inside the brackets, switch signs.

Example 2

  1. \displaystyle -(x+y) = (-1) \cdot (x+y) = (-1)x + (-1)y = -x-y
  2. \displaystyle -(x^2-x) = (-1) \cdot (x^2-x) = (-1)x^2 -(-1)x = -x^2 +x
    where we have in the final step used \displaystyle -(-1)x = (-1)(-1)x = 1\cdot x = x\,\mbox{.}
  3. \displaystyle -(x+y-y^3) = (-1)\cdot (x+y-y^3) = (-1)\cdot x + (-1) \cdot y -(-1)\cdot y^3
    \displaystyle \phantom{-(x+y-y^3)}{} = -x-y+y^3
  4. \displaystyle x^2 - 2x -(3x+2) = x^2 -2x -3x-2 = x^2 -(2+3)x -2
    \displaystyle \phantom{x^2-2x-(3x+2)}{} = x^2 -5x -2

If the distributive law is applied in backwards we say we factor the expression. One often would like to to factorise out as large a numerical factor as possible.

Example 3

  1. \displaystyle 3x +9y = 3x + 3\cdot 3y = 3(x+3y)
  2. \displaystyle xy + y^2 = xy + y\cdot y = y(x+y)
  3. \displaystyle 2x^2 -4x = 2x\cdot x - 2\cdot 2\cdot x = 2x(x-2)
  4. \displaystyle \frac{y-x}{x-y} = \frac{-(x-y)}{x-y} = \frac{-1}{1} = -1


Squaring rules

The distributive law occasionally has to be used repeatedly to deal with larger expression. If we consider

Vorlage:Fristående formel

and regard \displaystyle a+b as a factor that multiplies the bracketed expression(c+d) we get

Vorlage:Fristående formel

Then the \displaystyle c and the \displaystyle dare multiplied into their respective brackets,

Vorlage:Fristående formel

A mnemonic for this formula is:

2.1 - Figur - Distributiva lagen två gånger

Example 4

  1. \displaystyle (x+1)(x-2) = x\cdot x + x \cdot (-2) + 1 \cdot x + 1 \cdot (-2) = x^2 -2x+x-2
    \displaystyle \phantom{(x+1)(x-2)}{}=x^2 -x-2
  2. \displaystyle 3(x-y)(2x+1) = 3(x\cdot 2x + x\cdot 1 - y \cdot 2x - y \cdot 1) = 3(2x^2 +x-2xy-y)
    \displaystyle \phantom{3(x-y)(2x+1)}{}=6x^2 +3x-6xy-3y
  3. \displaystyle (1-x)(2-x) = 1\cdot 2 + 1 \cdot (-x) -x\cdot 2 - x\cdot (-x) = 2-x-2x+x^2
    \displaystyle \phantom{(1-x)(2-x)}{}=2-3x+x^2    where we have used \displaystyle -x\cdot (-x) = (-1)x \cdot (-1)x = (-1)^2 x^2 = 1\cdot x^2 = x^2.

Two important special cases of the above formula is when \displaystyle a+b and \displaystyle c+d are the same expression

Squaring rules Vorlage:Fristående formel Vorlage:Fristående formel

These formulas are called the first and second squaring rules

Example 5

  1. \displaystyle (x+2)^2 = x^2 + 2\cdot 2x+ 2^2 = x^2 +4x +4
  2. \displaystyle (-x+3)^2 = (-x)^2 + 2\cdot 3(-x) + 3^2 = x^2 -6x +9
    where \displaystyle (-x)^2 = ((-1)x)^2 = (-1)^2 x^2 = 1 \cdot x^2 = x^2\,\mbox{.}
  3. \displaystyle (x^2 -4)^2 = (x^2)^2 - 2 \cdot 4x^2 + 4^2 = x^4 -8x^2 +16
  4. \displaystyle (x+1)^2 - (x-1)^2 = (x^2 +2x +1)- (x^2-2x+1)
    \displaystyle \phantom{(x+1)^2-(x-1)^2}{}= x^2 +2x +1 -x^2 + 2x-1
    \displaystyle \phantom{(x+1)^2-(x-1)^2}{} = 2x+2x = 4x
  5. \displaystyle (2x+4)(x+2) = 2(x+2)(x+2) = 2(x+2)^2 = 2(x^2 + 4x+ 4)
    \displaystyle \phantom{(2x+4)(x+2)}{}=2x^2 + 8x + 8
  6. \displaystyle (x-2)^3 = (x-2)(x-2)^2 = (x-2)(x^2-4x+4)
    \displaystyle \phantom{(x-2)^3}{}=x \cdot x^2 + x\cdot (-4x) + x\cdot 4 - 2\cdot x^2 - 2 \cdot (-4x)-2 \cdot 4
    \displaystyle \phantom{(x-2)^3}{}=x^3 -4x^2 + 4x-2x^2 +8x -8 = x^3-6x^2 + 12x -8

The squaring rules are also used in the reverse direction to factorise expressions.

Example 6

  1. \displaystyle x^2 + 2x+ 1 = (x+1)^2
  2. \displaystyle x^6-4x^3 +4 = (x^3)^2 - 2\cdot 2x^3 +2^2 = (x^3-2)^2
  3. \displaystyle x^2 +x + \frac{1}{4} = x^2 + 2\cdot\frac{1}{2}x + \bigl(\frac{1}{2}\bigr)^2 = \bigl(x+\frac{1}{2}\bigr)^2


Difference of two squares

A third special case of the first formula in the last section is the difference of two squares rule.

Konjugatregeln: Vorlage:Fristående formel

This formula can be obtained directly by expanding the left hand side

Vorlage:Fristående formel

Example 7

  1. \displaystyle (x-4y)(x+4y) = x^2 -(4y)^2 = x^2 -16y^2
  2. \displaystyle (x^2+2x)(x^2-2x)= (x^2)^2 - (2x)^2 = x^4 -4x^2
  3. \displaystyle (y+3)(3-y)= (3+y)(3-y) = 3^2 -y^2 = 9-y^2
  4. \displaystyle x^4 -16 = (x^2)^2 -4^2 = (x^2+4)(x^2-4) = (x^2+4)(x^2-2^2)
    \displaystyle \phantom{x^4-16}{}=(x^2+4)(x+2)(x-2)


Rational expressions

Calculations of fractions containing algebraic expressions are largely similar to ordinary calculations with fractions.

Multiplication and division of fractions containing algebraic expressions follow the same rules that apply to ordinary fractions,

Vorlage:Fristående formel

Example 8

  1. \displaystyle \frac{3x}{x-y} \cdot \frac{4x}{2x+y} = \frac{3x\cdot 4x}{(x-y)\cdot(2x+y)} = \frac{12x^2}{(x-y)(2x+y)}
  2. \displaystyle \frac{\displaystyle \frac{a}{x}}{\displaystyle \frac{x+1}{a}} = \frac{a^2}{x(x+1)}
  3. \displaystyle \frac{\displaystyle \frac{x}{(x+1)^2}}{\displaystyle \frac{x-2}{x-1}} = \frac{x(x-1)}{(x-2)(x+1)^2}

A fractional expression can have its numerator and denominator multiplied by the same factor

Vorlage:Fristående formel

The opposite of this, is cancellation, where we delete factors that the numerator and denominator have in common Vorlage:Fristående formel

Example 9

  1. \displaystyle \frac{x}{x+1} = \frac{x}{x+1} \cdot \frac{x+2}{x+2} = \frac{x(x+2)}{(x+1)(x+2)}
  2. \displaystyle \frac{x^2 -1}{x(x^2-1)}= \frac{1}{x}
  3. \displaystyle \frac{(x^2-y^2)(x-2)}{(x^2-4)(x+y)} = \left\{\,\text{konjugatregeln}\,\right\} = \frac{(x+y)(x-y)(x-2)}{(x+2)(x-2)(x+y)} = \frac{x-y}{x+2}

When fractional expressions are added or subtracted, they may need to be converted so that they have the same denominator before the numerators can be combined together,


Vorlage:Fristående formel

One normally tries to convert the fractions by multiplying the numerators and denominators by minimal factors to facilitate the calculations. The lowest common denominator (LCD) is the common denominator which contains the least number of factors.

Example 10

  1. \displaystyle \frac{1}{x+1} + \frac{1}{x+2}\quad has \displaystyle \ \text{LCD} = (x+1)(x+2)

    Convert the first term using \displaystyle (x+2) and the second term using \displaystyle (x+1) Vorlage:Fristående formel
  2. \displaystyle \frac{1}{x} + \frac{1}{x^2}\quad has \displaystyle \ \text{LCD} = x^2

    We only need to convert the first term to get a common denominator Vorlage:Fristående formel
  3. \displaystyle \frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}\quad has \displaystyle \ \text{LCD}= x^2(x+1)^2(x+2)

    The first term is converted using \displaystyle x(x+2) while the other term is converted using \displaystyle (x+1)^2 Vorlage:Fristående formel
  4. \displaystyle \frac{x}{x+1} - \frac{1}{x(x-1)} -1 \quad has \displaystyle \ \text{LCD}=x(x-1)(x+1)

      We must convert all the terms so that they have the common denominator \displaystyle x(x-1)(x+1) Vorlage:Fristående formel

To simplify large expressions, it is often necessary to both cancel factors and multiply numerators and denominators by factors. As cancellation implies that we have performed factorisations, it is obvious we should try to keep expressions (such as the denominator) factorised and not expand something that we will later need to factorise.


Example 11

  1. \displaystyle \frac{1}{x-2} - \frac{4}{x^2-4} = \frac{1}{x-2} - \frac{4}{(x+2)(x-2)} = \left\{\,\mbox{MGN} = (x+2)(x-2)\,\right\}

    \displaystyle \phantom{\frac{1}{x-2} - \frac{4}{x^2-4}}{} = \frac{x+2}{(x+2)(x-2)} - \frac{4}{(x+2)(x-2)}

    \displaystyle \phantom{\frac{1}{x-2} - \frac{4}{x^2-4}}{} = \frac{x+2 -4}{(x+2)(x-2)} = \frac{x-2}{(x+2)(x-2)} = \frac{1}{x+2}
  2. \displaystyle \frac{x + \displaystyle \frac{1}{x}}{x^2+1} = \frac{\displaystyle \frac{x^2}{x} + \frac{1}{x}}{x^2+1} = \frac{\displaystyle \frac{x^2+1}{x}}{x^2+1} = \frac{x^2+1}{x(x^2+1)} = \frac{1}{x}
  3. \displaystyle \frac{\displaystyle \frac{1}{x^2} - \frac{1}{y^2}}{x+y} = \frac{\displaystyle \frac{y^2}{x^2y^2} - \frac{x^2}{x^2y^2}}{x+y} = \frac{\displaystyle \frac{y^2-x^2}{x^2y^2}}{x+y} = \frac{y^2-x^2}{x^2y^2(x+y)}

    \displaystyle \phantom{\smash{\frac{\displaystyle \frac{1}{x^2} - \frac{1}{y^2}}{x+y}}}{} = \frac{(y+x)(y-x)}{x^2y^2(x+y)} = \frac{y-x}{x^2y^2}


Exercises


Study advice

'The basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.


Keep in mind that:

Be careful. If you make a mistake somewhere the rest of the calculation will be wrong.

Use many intermediate steps . If you are unsure of a calculation do it in many small steps rather than one big step.

Do not expand unnecessarily. You later may be forced to faktorise what you earlier expanded.


Reviews

Learn more about algebra in the English Wikipedia

Understanding Algebra - English text on the Web


Länktips

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/2.1_Algebraische_Ausdr%C3%BCcke“