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Mathematik für Informatiker 3 (Vorlesung 2)

Skript-AnfangBlatt0 – Seite 1
Skript-EndeBlatt0 – Seite 1

Aufgabe 1

a)

\[ \frac{2\sqrt{15}}{\sqrt{6}} \]

\[ = \frac{2\sqrt{5}*\sqrt{3}}{\sqrt{3}*\sqrt{2}} \]

\[ = \frac{2\sqrt{5}}{\sqrt{2}} \]

\[ = 2*\sqrt{\frac{5}{2}} \]

\[ = \sqrt{\frac{4*5}{2}} \]

\[ = \sqrt{10} \]

b)

\[ \frac{6}{5-9} + \frac{7}{5+9} \]

\[ = \frac{6}{-4} + \frac{7}{14} \]

\[ = \frac{6}{-4} + \frac{1}{2} \]

\[ = -\frac{6}{4} + \frac{1}{2} \]

\[ = \frac{-6+2}{4} \]

\[ = \frac{-4}{4} \]

\[ = 1 \]

c)

\[\frac{\frac{3}{4}+\frac{1}{6}}{\frac{1}{2}+\frac{1}{6}} \]

\[ = \frac{\frac{18+4}{24}}{\frac{3-2}{6}} \]

\[ = \frac{\frac{22}{24}}{\frac{1}{6}} \]

\[ = \frac{22*6}{24*1} \]

\[ = \frac{22*1}{4} \]

\[ = \frac{11}{2} \]

d)

\[ \frac{-2*(4-7)}{\frac{1}{2}-\frac{7}{2}} \]

\[ = \frac{6}{-\frac{6}{2}} \]

\[ = \frac{6}{-3} \]

\[ = -2 \]

e)

\[ 3*(2-(-1))-(-2)(-1+3) \]

\[ = 3*(3)-(-2)(2) \]

\[ = 9 + 4 \]

\[ = 13 \]

f)

\[ (x^{2}-1)(x-x^{2})(-\frac{1}{x}) \]

\[ = (x^{2}-1)(-\frac{x}{x}+\frac{x^{2}}{x}) \]

\[ = (x^{2}-1)(-1+x) \]

\[ = -x^{2}+x^{3}+1-x \]

\[ = x^{3}-x^{2}-x+1 \]

g)

\[ \frac{a^{2}-4}{(a+2)^{2}} \]

\[ = \frac{(a+2)(a-2)}{(a+2)^{2}} \]

\[ = \frac{a-2}{a+2} \]

j)

\[ \frac{a-1}{a^{2}+a}-\frac{1}{a}+\frac{4}{a^{2}-1}-\frac{a+1}{a^{2}-a} \]

\[ = \frac{(a-1)}{a*(a+1)}-\frac{(a+1)(a-1)}{a*(a+1)(a-1)}+\frac{4}{a^{2}-1}-\frac{a+1}{a*(a-1)} \]

\[ = \frac{(a-1)(a-1)}{a*(a+1)(a-1))}-\frac{(a+1)(a-1)}{a*(a+1)(a-1)}+\frac{4}{(a+1)(a-1)}-\frac{(a+1)(a+1)}{a*(a-1)(a+1)} \]

\[ = \frac{(a-1)(a-1)-(a+1)(a-1)+4a-(a+1)(a+1)}{a(a+1)(a-1)} \]

\[ = \frac{a^{2}-2a+1-a^{2}+1+4a-a^{2}-2a-1}{a(a+1)(a-1)} \]

\[ = \frac{-a^{2}-1}{a(a+1)(a-1)} \]

\[ = \frac{-(a^{2}+1)}{a(a+1)(a-1)} \]

\[ = \frac{-(a+1)(a-1)}{a(a+1)(a-1)} \]

\[ = -\frac{1}{a} \]

i)

\[ \frac{\frac{x+1}{x-1}-1}{\frac{x+1}{x-1}+1} \]

\[ = \frac{\frac{x+1}{x-1}-\frac{x-1}{x-1}}{\frac{x+1}{x-1}+\frac{x-1}{x-1}} \]

\[ = \frac{\frac{x+1-x+1}{x-1}}{\frac{x+1+x-1}{x-1}} \]

\[ = \frac{\frac{2}{x-1}}{\frac{2x}{x-1}} \]

\[ = \frac{2*(x-1)}{(x-1)*2x} \]

\[ = \frac{2}{2x} \]

\[ = \frac{1}{x} \]

l)

\[ 4a^{2}b – 5ab^{2} + 2a^{2}b + 11ab^{2} \]

\[ = 6a^{2}b + 6ab^{2} \]

\[ = 6(a^{2}b+ab^{2}) \]

\[ = 6ab(a+b) \]

k)

\[ \frac{\frac{1}{a^{2}}-\frac{2}{ab}+\frac{1}{b^{2}}}{\frac{1}{a^{2}}-\frac{1}{b^{2}}} \]

\[ = \frac{(\frac{1}{a}-\frac{1}{b})^{2}}{(\frac{1}{a}+\frac{1}{b})(\frac{1}{a}-\frac{1}{b})} \]

\[ = \frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}} \]

\[ = \frac{\frac{b}{ab}-\frac{a}{ab}}{\frac{b}{ab}+\frac{a}{ab}} \]

\[ = \frac{\frac{b-a}{ab}}{\frac{b+a}{ab}} \]

\[ = \frac{(b-a)}{ab}*\frac{ab}{(b+a)} \]

\[ = \frac{b-a}{b+a} \]

oder:

\[ \frac{\frac{1}{a^{2}}-\frac{2}{ab}+\frac{1}{b^{2}}}{\frac{1}{a^{2}}-\frac{1}{b^{2}}} \]

\[ = \frac{\frac{b^{2}}{a^{2}b^{2}}-\frac{2ab}{a^{2}b^{2}}+\frac{a^{2}}{a^{2}b^{2}}}{\frac{b^{2}-a^{2}}{a^{2}b^{2}}} \]

\[ = \frac{(\frac{1}{a^{2}b^{2}})(b^{2}-2ab+a^{2})}{(\frac{1}{a^{2}b^{2}})(b^{2}-a^{2})} \]

\[ = \frac{b^{2}-2ab+a^{2}}{b^{2}-a^{2}} \]

\[ = \frac{(b-a)^{2}}{(b-a)(b+a)} \]

\[ = \frac{b-a}{b+a} \]

Aufgabe 2

a)

\[ (-1,3) \cap [1,5] = [1,3) \]

b)

\[ (-1,3) \cup [1,5] = (-1,5] \]

c)

\[ (-1,3) \setminus [1,5] = (-1,1) \]

d)

\[ \{x \in \mathbb{R}:x^{2}<4\} = (-2,2) \]

e

\[ \{x \in \mathbb{R}:|x-2|\leq 3\} = [-1,5] \]

f)

\[ \{x \in \mathbb{R}:x+1\geq 3-x\} = [1,\infty ] \]

g)

\[ \{x \in \mathbb{R}:(x+2)(x-1)<0\} = (-2,1) \]

Aufgabe 3

b)

\[ \log_2(\frac{\sqrt{x^{2}*x^{3}}^{3}}{x^{5}\sqrt{x}\sqrt[3]{x}}) = 1+\frac{2}{3}\log_2x \]

\[ \log_2(\sqrt{x^{2}*x^{3}}^{3}) – \log_2(x^{5}\sqrt{x}\sqrt[3]{x}) = 1+\frac{2}{3}\log_2x \]

\[ 3\log_2(\sqrt{x^{2}*x^{3}}) – \log_2(x^{5}\sqrt{x}\sqrt[3]{x}) = 1+\frac{2}{3}\log_2x \]

\[ 3\log_2(\sqrt{x^{2}*x^{3}}) – (\log_2(x^{5})+\log_2(\sqrt{x})+\log_2(\sqrt[3]{x})) = 1+\frac{2}{3}\log_2x \]

\[ \frac{3}{2}\log_2(x^{5}) – (5\log_2(x)+\frac{1}{2}\log_2(x)+\frac{1}{3}\log_2(x)) = 1+\frac{2}{3}\log_2x \]

\[ \frac{15}{2}\log_2(x) – 5\log_2(x)-\frac{1}{2}\log_2(x)-\frac{1}{3}\log_2(x) = 1+\frac{2}{3}\log_2x \]

\[ \log_2x(\frac{15}{2}-5-\frac{1}{2}-\frac{1}{3}) = 1+\frac{2}{3}\log_2x \]

\[ (\frac{5}{3})\log_2x = 1+\frac{2}{3}\log_2x \]

\[ \log_2x = 1 \]

\[ 2^{1} = x \]

\[ L ={2} \]

c)

\[ (x+1)^{2} \geq 4 \]

\[ (x+1) \geq \sqrt{4} \wedge (x+1) \leq -\sqrt{4} \]

\[ (x+1) \geq 2 \wedge (x+1) \leq -2 \]

\[ x \geq 1 \wedge x \leq -3 \]

\[ L = (-\infty ;-3] \cup [1;+\infty) \]

\[ L = \mathbb{R}\setminus (-3,1) \]

Aufgabe 4

a)

\[ \sqrt{x+2}-1 \geq 0 \]

\[ \sqrt{x+2} \geq 1 \]

\[ x+2 \geq 1 \]

\[ x \geq -1 \]

\[ x+2 \geq 0 \]

\[ x \geq -2 \]

\[ y=f(x) für x \in [-1,\infty ] \]

b)

\[ e^{y}= \sqrt{x+2}-1 \]

\[ e^{y} +1 = \sqrt{x+2} \]

\[ (e^{y}+1)^{2} = x+2 \]

\[ e^{2y}+2e^{y}+1 = x+2 \]

\[ x = f^{-1}(y) = e^{2y}+2e{y}-1 \]

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