solvefor x,(f-g)(x)=((x^2+2x-1))/x

Lösung

löse nach

x, (f - g) (x) = \frac{( x ^{2} + 2 x - 1 )}{x}

x = \frac{- 1 + g + 2 - f}{1 - f + g}, x = - \frac{g + 2 - f + 1}{g + 1 - f}; 1 - f + g \neq = 0

Schritte zur Lösung

(f - g) (x) = \frac{( x ^{2} + 2 x - 1 )}{x}

Vereinfache

\frac{( x ^{2} + 2 x - 1 )}{x} : \frac{x ^{2} + 2 x - 1}{x}

(f - g) x = \frac{x ^{2} + 2 x - 1}{x}

Multipliziere beide Seiten mit

x

x^{2} (f - g) = x^{2} + 2 x - 1

Löse

x^{2} (f - g) = x^{2} + 2 x - 1 : x = \frac{- 1 + g + 2 - f}{1 - f + g}, x = - \frac{g + 2 - f + 1}{g + 1 - f}; 1 - f + g \neq = 0

x = \frac{- 1 + g + 2 - f}{1 - f + g}, x = - \frac{g + 2 - f + 1}{g + 1 - f}; 1 - f + g \neq = 0

\frac{1}{x - 2} + \frac{3}{x + 3} = 6

63 = \frac{3 \times ( x ^{3} - 1 )}{x - 1}

8 + \frac{6}{x} = x + 9

8 = \frac{16}{x}

\frac{3 x - 1}{x} = \frac{5}{2}