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Version 5.1 von Kim Fujan am 2025/05/20 13:10
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1 (% class="abc" %)
2 1. {{formula}} x^{2}+2x-3=0 {{/formula}}
3
4 Lösung mit abc-Formel:
5 {{formula}}x_{1,2}=\frac{2\pm \sqrt{4+12}}{2}=\frac{2\pm 4}{2}{{/formula}}
6 {{formula}} x_{1}=3 \quad ; \quad x_{2}=-1 {{/formula}}
7
8 1. {{formula}} e^{2x}+2e^x-3=0 {{/formula}}
9
10 Substitution: {{formula}} e^x=u {{/formula}}
11 {{formula}} u^{2}+2u-3=0 {{/formula}}
12
13 Lösung mit abc-Formel:
14 {{formula}}u_{1,2}=\frac{2\pm \sqrt{4+12}}{2}=\frac{2\pm 4}{2}{{/formula}}
15 {{formula}} u_{1}=3 \quad ; \quad u_{2}=-1 {{/formula}}
16
17 Resubstitution:
18 {{formula}} e^x=3 \quad \Longleftrightarrow \quad x=ln(3) {{/formula}}
19 {{formula}} e^x=-1 \quad \Longleftrightarrow \quad {{/formula}} keine weitere Lösung!
20
21
22 1. {{formula}} e^x+2e^{\frac{1}{2}x}-3=0 {{/formula}}
23
24 Substitution: {{formula}} e^{\frac{1}{2}x}=u {{/formula}}
25 {{formula}} u^{2}+2u-3=0 {{/formula}}
26
27 Lösung mit abc-Formel:
28 {{formula}}u_{1,2}=\frac{2\pm \sqrt{4+12}}{2}=\frac{2\pm 4}{2}{{/formula}}
29 {{formula}} u_{1}=3 \quad ; \quad u_{2}=-1 {{/formula}}
30
31 Resubstitution:
32 {{formula}} e^{\frac{1}{2}x}=3 \quad \Longleftrightarrow \quad x=2 \cdot ln(3) {{/formula}}
33 {{formula}} e^{\frac{1}{2}x}=-1 \quad \Longleftrightarrow \quad {{/formula}} keine weitere Lösung!
34
35 1. {{formula}} e^x-2-\frac{15}{e^x}}=0 {{/formula}}
36
37 1. {{formula}} 2e^{4x}=e^{2x}+3 {{/formula}}