erweitern (x-2)(x+2)[x-2/(x^2)-4+1/x+2= 1/x-2]

Lösung

erweitern

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

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

Schritte zur Lösung

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

Multipliziere aus

(x - 2) (x + 2) : x^{2} - 4

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

Multipliziere aus

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

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

e x p an d \frac{1 + ln ( x )}{x ^{2} ( 1 - ln ( x ) ) ^{3}}

e x p an d \frac{( x - 3 ) ( x + 6 )}{x + 7}

e x p an d y^{'} (x) + \frac{( 1 - x ) y}{x}

e x p an d (y^{- 1})^{4}

e x p an d - \frac{( 6 x - 5 )}{( 3 x ^{2} - 5 x + 7 ) ^{2}}