erweitern (t(x_{2}-x_{1})(2+x_{1}+x_{2}))/3

Lösung

erweitern

\frac{t ( x _{2} - x _{1} ) ( 2 + x _{1} + x _{2} )}{3}

\frac{2 t x _{2}}{3} + \frac{t x _{2}^{2}}{3} - \frac{2 t x _{1}}{3} - \frac{t x _{1}^{2}}{3}

Schritte zur Lösung

\frac{t ( x _{2} - x _{1} ) ( 2 + x _{1} + x _{2} )}{3}

Multipliziere aus

t (x_{2} - x_{1}) (2 + x_{1} + x_{2}) : 2 t x_{2} + t x_{2}^{2} - 2 t x_{1} - t x_{1}^{2}

= \frac{2 t x _{2} + t x _{2}^{2} - 2 t x _{1} - t x _{1}^{2}}{3}

Wende Bruchregel an:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2 t x _{2} + t x _{2}^{2} - 2 t x _{1} - t x _{1}^{2}}{3} = \frac{2 t x _{2}}{3} + \frac{t x _{2}^{2}}{3} - \frac{2 t x _{1}}{3} - \frac{t x _{1}^{2}}{3}

= \frac{2 t x _{2}}{3} + \frac{t x _{2}^{2}}{3} - \frac{2 t x _{1}}{3} - \frac{t x _{1}^{2}}{3}

e x p an d \frac{3 ( 15 x - 3 )}{( x + 2 )}

e x p an d \frac{12 e ^{- 2 s} + 12 e ^{- 5 s} + 3 s}{s ( 3 + s )}

e x p an d (2 x + e^{x}) (x^{2} - e^{x})

e x p an d \frac{7 ( e - 1 )}{1 - e}

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