展开 (1-t,3(1-t)(4+t),-(1-t+2-t))(-1,1,-1)

解答

展开

(1 - t, 3 (1 - t) (4 + t), - (1 - t + 2 - t)) (- 1, 1, - 1)

- 1 \cdot 1, 1, - 1 \cdot 1 + t, 3, 1, 1, t, 3, 1 \cdot t^{4} + t, 3, 1, 1, t, 3, 1 \cdot 6 t^{3} + t, 3, 1, 1, t, 3, 1 \cdot t^{2} - t, 3, 1, 1, t, 3, 1 \cdot 24 t + t, 3, 1, 1, t, 3, 1 \cdot 16 + 1, 1, 1 \cdot 4 t^{2} - 1, 1, 1 \cdot 12 t + 1, 1, 1 \cdot 9

求解步骤

(1 - t, 3 (1 - t) (4 + t), - (1 - t + 2 - t)) (- 1, 1, - 1)

= (- 1 \cdot 1,, - 1) (1 - 3,, t (1 - t) (4 + t) - (1 - t + 2 - t))

乘开

(1 - t, 3 (1 - t) (4 + t), - (1 - t + 2 - t)) (- 1, 1, - 1) : - 1 \cdot 1, 1, - 1 \cdot 1 + t, 3 (1 - t) (4 + t), 1, 1, t, 3 (1 - t) (4 + t), 1 + 1 \cdot (1 - t + 2 - t), 1, 1 \cdot (1 - t + 2 - t)

= - 1 \cdot 1, 1, - 1 \cdot 1 + t, 3 (1 - t) (4 + t), 1, 1, t, 3 (1 - t) (4 + t), 1 + 1 \cdot (1 - t + 2 - t), 1, 1 \cdot (1 - t + 2 - t)

乘开

t, 3 (1 - t) (4 + t), 1, 1, t, 3 (1 - t) (4 + t), 1 : t, 3, 1, 1, t, 3, 1 \cdot t^{4} + t, 3, 1, 1, t, 3, 1 \cdot 6 t^{3} + t, 3, 1, 1, t, 3, 1 \cdot t^{2} - t, 3, 1, 1, t, 3, 1 \cdot 24 t + t, 3, 1, 1, t, 3, 1 \cdot 16

= - 1 \cdot 1, 1, - 1 \cdot 1 + t, 3, 1, 1, t, 3, 1 \cdot t^{4} + t, 3, 1, 1, t, 3, 1 \cdot 6 t^{3} + t, 3, 1, 1, t, 3, 1 \cdot t^{2} - t, 3, 1, 1, t, 3, 1 \cdot 24 t + t, 3, 1, 1, t, 3, 1 \cdot 16 + 1 \cdot (1 - t + 2 - t), 1, 1 \cdot (1 - t + 2 - t)

乘开

1 \cdot (1 - t + 2 - t), 1, 1 \cdot (1 - t + 2 - t) : 1, 1, 1 \cdot 4 t^{2} - 1, 1, 1 \cdot 12 t + 1, 1, 1 \cdot 9

= - 1 \cdot 1, 1, - 1 \cdot 1 + t, 3, 1, 1, t, 3, 1 \cdot t^{4} + t, 3, 1, 1, t, 3, 1 \cdot 6 t^{3} + t, 3, 1, 1, t, 3, 1 \cdot t^{2} - t, 3, 1, 1, t, 3, 1 \cdot 24 t + t, 3, 1, 1, t, 3, 1 \cdot 16 + 1, 1, 1 \cdot 4 t^{2} - 1, 1, 1 \cdot 12 t + 1, 1, 1 \cdot 9

e x p an d \frac{2 x + 1}{U ^{6} ( x - 3 ) ^{6} x}

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

e x p an d (f (x) = 2, x + 5) \cdot (g (x) = x + 5)

e x p an d - 3 (3 x - 5 = 7 - 4 x)

e x p an d (4 + 6 x) x^{3} e^{6 x}