Lösung 2.1:2a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
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,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
(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)\\
Zeile 9: Zeile 9:
Then, gather together ''x''²-, ''x''- and the constant terms and simplify
Then, gather together ''x''²-, ''x''- and the constant terms and simplify
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\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\\
&= \rlap{-5x^2+20\,\textrm{.}}\phantom{x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)}
&= \rlap{-5x^2+20\,\textrm{.}}\phantom{x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)}
\end{align}</math>}}
\end{align}</math>}}

Version vom 08:21, 22. Okt. 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,

\displaystyle \begin{align}

(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\,\textrm{.} \end{align}

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

\displaystyle \begin{align}

\phantom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\ &= -5x^2+0+20\\ &= \rlap{-5x^2+20\,\textrm{.}}\phantom{x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)} \end{align}

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