2.1 Algebraische Ausdrücke

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

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

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

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.

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

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

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)

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}

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{Difference of two squares}\,\right\} = \frac{(x+y)(x-y)(x-2)}{(x+2)(x-2)(x+y)} = \frac{x-y}{x+2}

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:Displayed math
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:Displayed math
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:Displayed math
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:Displayed math

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}