Lösung 2.1:2a

Aus Online Mathematik Brückenkurs 1

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First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,
First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,
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<math>
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{{Displayed math||<math>\begin{align}
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\qquad
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\begin{align}
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(x-4)(x-5)-3x(2x-3)&= x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)\\
(x-4)(x-5)-3x(2x-3)&= x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)\\
&= x^2-5x-4x+20-(6x^2-9x)\\
&= x^2-5x-4x+20-(6x^2-9x)\\
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&=x^2-5x-4x+20-6x^2+9x
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&=x^2-5x-4x+20-6x^2+9x\,\textrm{.}
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\end{align}
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\end{align}</math>}}
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</math>
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Then, gather together <math>x^2-, \, x- </math> and the constant terms and simplify
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Then, gather together ''x''²-, ''x''- and the constant terms and simplify
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<math>
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{{Displayed math||<math>\begin{align}
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\qquad
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\begin{align}
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\phantom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\
\phantom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\
&= -5x^2+0+20\\
&= -5x^2+0+20\\
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&= -5x^2+20
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&= \rlap{-5x^2+20\,\textrm{.}}\phantom{x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)}
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\end{align}
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\end{align}</math>}}
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</math>
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Version vom 08:10, 23. Sep. 2008

First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,

Vorlage:Displayed math

Then, gather together x²-, x- and the constant terms and simplify

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:2a“