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a/(a+b)+a*a/(a*a-b*b)+b/(a-b)=

 21.10.2015

Beste Antwort 

 #2
avatar+26256 
+34

a/(a+b)+a*a/(a*a-b*b)+b/(a-b)=?

 

\(\begin{array}{lcl} \frac{a}{a+b} +a\cdot \left( \frac{a}{a\cdot a-b\cdot b} \right) +\frac{b}{a-b}&& \\\\ =\frac{a}{a+b} +\frac{a\cdot a}{a\cdot a-b\cdot b} +\frac{b}{a-b} \\\\ =\frac{a}{a+b} +\frac{a^2}{a^2-b^2} +\frac{b}{a-b} \qquad | \qquad \text{3. Binom }~ a^2-b^2 = (a+b)(a-b)\\\\ =\frac{a}{a+b} +\frac{a^2}{(a+b)(a-b)} +\frac{b}{a-b} \\\\ =\frac{a}{a+b} +\frac{a}{a+b} \cdot \frac{a}{a-b} +\frac{b}{a-b} \\\\ =\frac{a}{a+b} \left( 1+ \frac{a}{a-b} \right) +\frac{b}{a-b} \\\\ =\frac{a}{a+b} \left( \frac{a-b+a}{a-b} \right) +\frac{b}{a-b} \\\\ =\frac{a}{a+b} \left( \frac{2a-b}{a-b} \right) +\frac{b}{a-b} \\\\ =\frac{1}{a-b} \left( \frac{a(2a-b)}{a+b} +b \right) \\\\ =\frac{1}{a-b} \left( \frac{a(2a-b)+b(a+b)}{a+b} \right) \\\\ =\frac{a(2a-b)+b(a+b)}{(a-b)(a+b)}\\\\ =\frac{2a^2-ab+ba+b^2}{(a-b)(a+b)} \qquad | \qquad -ab+ba = 0\\\\ =\frac{2a^2+b^2}{(a-b)(a+b)} \qquad | \qquad \text{3. Binom }~ (a+b)(a-b)= a^2-b^2 \\\\ \mathbf{=\frac{2a^2+b^2}{a^2-b^2 }} \\\\ \mathbf{\frac{a}{a+b} +a\cdot \left( \frac{a}{a\cdot a-b\cdot b} \right) +\frac{b}{a-b}=\frac{2a^2+b^2}{a^2-b^2 }} \\\\ \end{array}\)

laugh

 22.10.2015
 #1
avatar+12514 
+4

Hier kommt Dein Lösungsweg:

laugh

 21.10.2015
 #2
avatar+26256 
+34
Beste Antwort

a/(a+b)+a*a/(a*a-b*b)+b/(a-b)=?

 

\(\begin{array}{lcl} \frac{a}{a+b} +a\cdot \left( \frac{a}{a\cdot a-b\cdot b} \right) +\frac{b}{a-b}&& \\\\ =\frac{a}{a+b} +\frac{a\cdot a}{a\cdot a-b\cdot b} +\frac{b}{a-b} \\\\ =\frac{a}{a+b} +\frac{a^2}{a^2-b^2} +\frac{b}{a-b} \qquad | \qquad \text{3. Binom }~ a^2-b^2 = (a+b)(a-b)\\\\ =\frac{a}{a+b} +\frac{a^2}{(a+b)(a-b)} +\frac{b}{a-b} \\\\ =\frac{a}{a+b} +\frac{a}{a+b} \cdot \frac{a}{a-b} +\frac{b}{a-b} \\\\ =\frac{a}{a+b} \left( 1+ \frac{a}{a-b} \right) +\frac{b}{a-b} \\\\ =\frac{a}{a+b} \left( \frac{a-b+a}{a-b} \right) +\frac{b}{a-b} \\\\ =\frac{a}{a+b} \left( \frac{2a-b}{a-b} \right) +\frac{b}{a-b} \\\\ =\frac{1}{a-b} \left( \frac{a(2a-b)}{a+b} +b \right) \\\\ =\frac{1}{a-b} \left( \frac{a(2a-b)+b(a+b)}{a+b} \right) \\\\ =\frac{a(2a-b)+b(a+b)}{(a-b)(a+b)}\\\\ =\frac{2a^2-ab+ba+b^2}{(a-b)(a+b)} \qquad | \qquad -ab+ba = 0\\\\ =\frac{2a^2+b^2}{(a-b)(a+b)} \qquad | \qquad \text{3. Binom }~ (a+b)(a-b)= a^2-b^2 \\\\ \mathbf{=\frac{2a^2+b^2}{a^2-b^2 }} \\\\ \mathbf{\frac{a}{a+b} +a\cdot \left( \frac{a}{a\cdot a-b\cdot b} \right) +\frac{b}{a-b}=\frac{2a^2+b^2}{a^2-b^2 }} \\\\ \end{array}\)

laugh

heureka 22.10.2015

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