Algebra-Terme-Binomische Formel
$(a + b)^{2} $
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
$ (a - b)^{2}$
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
$(a + b)\cdot (a - b)$
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
$(ax+b)^3$
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
$(ax+b)^4$
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Beispiel Nr: 09
$\begin{array}{l} (ax+b)^4 =a^4\cdot x^4 +4a^3x^3b +4a^2x^2b^2 +4axb^3 +b^4 \\ (ax+b)^4\\ \textbf{Gegeben:} \\ (1\frac{1}{4}x + \frac{4}{5})^{4}\\ \\ \textbf{Rechnung:} \\
\\(1\frac{1}{4}x+\frac{4}{5})^{4}=\left(1\frac{1}{4}\right)^{4}x^{4}+4 \cdot \left(1\frac{1}{4}\right)^3\cdot x^3\cdot \frac{4}{5}+6 \cdot \left(1\frac{1}{4}\right)^2\cdot x^2\cdot \left(\frac{4}{5}\right)^2+4\cdot 1\frac{1}{4}\cdot x\cdot \left(\frac{4}{5}\right)^3+\left(\frac{4}{5}\right)^{4}
\\(1\frac{1}{4}x+\frac{4}{5})^{4}=2,44x^4+6\frac{1}{4}x^3+6x^2+2\frac{14}{25}x+0,41
\end{array}$