Wiki-Quellcode von Lösung Termumformungen

Version 4.1 von akukin am 2025/08/15 14:27

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1 Vereinfache:
2 1.a)
3
4 {{formula}}
5 \begin{align*}
6 &\color{blue}{2(4a - 5) - 3(2a - 3) + 4(-3a + 5)} \\
7 &= 8a - 10 - 6a + 9 - 12a + 20 = \textbf{-10a + 19}
8 \end{align*}
9 {{/formula}}
10
11 1.b)
12
13 {{formula}}
14 \begin{align*}
15 &\color{blue}{x - (x + 3) - 4(-x + 1)}\\
16 &= x - x - 3 + 4x - 4 = \textbf{4x - 7}
17 \end{align*}
18 {{/formula}}
19
20 2.a)
21
22 {{formula}}
23 \begin{align*}
24 &\color{blue}{6a - 2(7b - (4a + 3b)) + 2((2a - b) - 7a)}\\
25 &= 6a - 2(7b - 4a - 3b) + 2(2a - b - 7a) \\
26 &= 6a - 14b + 8a + 6b + 4a - 2b - 14a = \textbf{4a - 10b}
27 \end{align*}
28 {{/formula}}
29
30 2.b)
31
32 {{formula}}
33 \begin{align*}
34 &\color{blue}{2x + 3(4 - (2x + 1) + 3x)}\\
35 &= 2x + 3(4 - 2x - 1 + 3x)\\
36 &= 2x + 3(3 + x) = 2x + 9 + 3x = \textbf{5x + 9}
37 \end{align*}
38 {{/formula}}
39
40 Multipliziere aus:
41
42 3.a) {{formula}}\color{blue}{(3a + b)(a - 5b)} = \mathbf{3a^2 - 14ab - 5b^2}{{/formula}}
43 3.b) {{formula}}(4x - 3)(-x + \frac{1}{3})= \mathbf{-4x^2 + \frac{13}{3}x - 1}{{/formula}}
44
45 4.a) {{formula}}\color{blue}{(2x + y)^2}= \mathbf{4x^2 + 4xy + y^2}{{/formula}}
46 4.b) {{formula}}\color{blue}{(x - 3y)^2}= \mathbf{x^2 - 6xy + 9y^2}{{/formula}}
47 4.c) {{formula}}\color{blue}{(x^2 - 2)(x^2 + 2)}= \mathbf{x^4 - 4}{{/formula}}
48 4.d)
49
50 {{formula}}
51 \begin{align*}
52 &\color{blue}{(3 - x)^2 - (x + 1)^2 + 2(x - 1)(x + 1)}\\
53 &= (9 - 6x + x^2) - (x^2 + 2x + 1) + 2(x^2 - 1)\\
54 &= 9 - 6x + x^2 - x^2 - 2x - 1 + 2x^2 - 2 = \mathbf{2x^2 - 8x + 6}
55 \end{align*}
56 {{/formula}}
57
58 Faktorisiere:
59
60 5.a) {{formula}}\color{blue}{12ax^2 - 8ax}= \mathbf{4ax(3x - 2)}{{/formula}}
61 5.b) {{formula}}\color{blue}{3x^2 - 12}= 3(x^2 - 4) = \mathbf{3(x - 2)(x + 2)}{{/formula}}
62 5.c) {{formula}}\color{blue}{\frac{3ax^2 - 3a}{9x + 9}}= \frac{3a(x^2 - 1)}{9(x + 1)} = \frac{a(x - 1)(x + 1)}{3(x + 1)} = \mathbf{\frac{a(x - 1)}{3}}{{/formula}}