a)ab−ac+dc−db=a(b+c)−d(b−c)=(b−c)(a−d)
b)ac+ad−bc−bd=a(c+d)−b(c+d)=(c+d)(a−b)
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a)ab−ac+dc−db=a(b+c)−d(b−c)=(b−c)(a−d)
b)ac+ad−bc−bd=a(c+d)−b(c+d)=(c+d)(a−b)