化简 (t-1/2)(t^2-t+1/6)

解答

化简

(t - \frac{1}{2}) (t^{2} - t + \frac{1}{6})

\frac{( 2 t - 1 ) ( 6 t ^{2} - 6 t + 1 )}{12}

求解步骤

(t - \frac{1}{2}) (t^{2} - t + \frac{1}{6})

化简

t - \frac{1}{2} : \frac{2 t - 1}{2}

= \frac{2 t - 1}{2} (t^{2} - t + \frac{1}{6})

使用分式法则:

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

= \frac{( 2 t - 1 ) ( t ^{2} - t + \frac{1}{6} )}{2}

化简

(2 t - 1) (t^{2} - t + \frac{1}{6}) : \frac{( 2 t - 1 ) ( t ^{2} \cdot 6 - t \cdot 6 + 1 )}{6}

= \frac{\frac{( 2 t - 1 ) ( t ^{2} \cdot 6 - t \cdot 6 + 1 )}{6}}{2}

使用分式法则:

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

= \frac{( 2 t - 1 ) ( t ^{2} \cdot 6 - t \cdot 6 + 1 )}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{( 2 t - 1 ) ( t ^{2} \cdot 6 - t \cdot 6 + 1 )}{12}

= \frac{( 2 t - 1 ) ( 6 t ^{2} - 6 t + 1 )}{12}

s im pl i f y \frac{1}{( x + y i ) ^{2} - 4}

e x p an d - cos (v) sin (u) (2 cos (v) + cos (v) cos (u))

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

e x p an d (sin (2 t) + 2 sin (t)) \cdot (2 sin (t) - 2 sin (2 t))

e x p an d \frac{s + 1}{( s + 30 ) ( s ^{2} + s + 1 ) ( s + 15 )}