LÖSUNG: Aufgaben zur Umformung von Gleichungen

Aufgabe

$\frac{a \cdot b}{c \cdot d} = e$ $a = \frac{ecd}{b}$ $b = \frac{ecd}{a}$ $c = \frac{ab}{ed}$ $d = \frac{ab}{ce}$ $e = \frac{ab}{cd}$

$a + b - c = d - e$ a=d-e-b+c b=d-e-a+c c=a+b-d+e d=a+b-c+e e=d-a-b+c

$\frac{a + b}{c - d} + e = \frac{1}{f}$ $a= \frac{1}{f} - e) (c - d) - b$ $b = \frac{1}{f} - e) (c - d) - a$ $c = \frac{a + b}{\frac{1}{f} - e} + d$ $d = c - \frac{a + b}{\frac{1}{f} - e}$ $e = \frac{1}{f} - \frac{a + b}{c - d}$ $f = \frac{1}{\frac{a + b}{c - d} + e}$

$a^{2} * b^{3} = \sqrt{c} + 2^{d} - \sin (e)$ $a = \sqrt{\frac{\sqrt{c} + 2^{d} - \sin e}{b^{3}}}$ $b = \sqrt[3]{\frac{\sqrt{c} + 2^{d} - \sin e}{a^{2}}}$ ${c = (a^{2} b^{2} - 2^{d} + \sin e)}^{2}$ $d = \ln (a^{2} b^{3} - \sqrt{c} + \sin e) / \ln 2$ $e = \arcsin (\sqrt{c} + 2^{d} - a^{2} b^{3})$

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Aufgabe
$\frac{a \cdot b}{c \cdot d} = e$	$a = \frac{ecd}{b}$	$b = \frac{ecd}{a}$	$c = \frac{ab}{ed}$	$d = \frac{ab}{ce}$	$e = \frac{ab}{cd}$
$a + b - c = d - e$	a=d-e-b+c	b=d-e-a+c	c=a+b-d+e	d=a+b-c+e	e=d-a-b+c
$\frac{a + b}{c - d} + e = \frac{1}{f}$	$a= \frac{1}{f} - e) (c - d) - b$	$b = \frac{1}{f} - e) (c - d) - a$	$c = \frac{a + b}{\frac{1}{f} - e} + d$	$d = c - \frac{a + b}{\frac{1}{f} - e}$	$e = \frac{1}{f} - \frac{a + b}{c - d}$	$f = \frac{1}{\frac{a + b}{c - d} + e}$
$a^{2} * b^{3} = \sqrt{c} + 2^{d} - \sin (e)$	$a = \sqrt{\frac{\sqrt{c} + 2^{d} - \sin e}{b^{3}}}$	$b = \sqrt[3]{\frac{\sqrt{c} + 2^{d} - \sin e}{a^{2}}}$	${c = (a^{2} b^{2} - 2^{d} + \sin e)}^{2}$	$d = \ln (a^{2} b^{3} - \sqrt{c} + \sin e) / \ln 2$	$e = \arcsin (\sqrt{c} + 2^{d} - a^{2} b^{3})$