超急！請幫忙這四道中一數🙏🏻（只懂一題也可，只要快！pls）?

(m+m) ÷ (n-n-n-n)  = (2m) ÷ (-2n) = -(m/n)

k ×(2k+k) × (2k-k) = k × (3k) × k = 3k^3

3q × q × 4p × p = (3q × q) × (4p×p) = 3q^2 × 4p^2 = 12 p^2 q^2

6s × 4s ÷ 3 × (-2t) =24s^2 ÷ 3 ×(-2t) = 8s^2 × (-2t) = -16 s^2 t


詳細:
(m+m) ÷ (n-n-n-n)
     = (2m) ÷ (-2n)   (m+m = 2m, n-n-n-n = -2n)
     = -(m/n)             (負號提出, 分子分母約掉共同因子2)

k ×(2k+k) × (2k-k)
    = k × (3k) × k   (2k+k = 3k, 2k-k = k)
    = (3k) × k × k   (交換律)
    = 3k^3               (結合律)

3q × q × 4p × p
    = (3q × q) × (4p×p)    (結合律)
    = 3q^2 × 4p^2            (3q × q = 3q^2, 4p × p = 4p^2)
    = 3 × 4 × q^2 × p^2    (交換律)
    = 12 p^2 q^2

6s × 4s ÷ 3 × (-2t) 
     =24s^2 ÷ 3 ×(-2t)    (6s × 4s = 24s^2)
    = 8s^2 × (-2t)           (24s^2 ÷ 3 = 8s^2)  
    = 8 × (-2) × s^2 × t   (交換律)         
    = -16 s^2 t