Question

1.Phân tích đa thức thành nhân tử:

a.3a²c²+bd+3abc+acd

b.ab\(\times\)(a+b)-bc\(\times\)(a+c)+abc

c.a\(\times\)(b²+c²)+b\(\times\)(c²+a²)+c\(\times\)(a²+b²)+2abc

d.bc\(\times\)(b+c)+ac\(\times\)(c-a)-ab\(\times\)(a+b)

Answer 1

3a²c² + bd + 3abc + acd

= 3ac(ac + b) + d(ac + b)

= (ac + b)(3ac + d)

ab(a + b) - bc(a + c) + abc

= b(a² + ab - ac - c² + ac)

= b(a² + ab - c²)

a(b² + c²) + b(c² + a²) + c(a² + b²) + 2abc

= ab² + ac² + bc² + a²b + c(a² + 2ab + b²)

= c²(a + b) + ab(a + b) + c(a + b)²

= (a + b)(c² + ab + ac + bc)

= (a + b)[c(b + c) + a(b + c)]

= (a + b)(a + c)(b + c)

bc(b + c) + ac(c - a) - ab(a + b)

= bc(b + c) + ac[(b + c) - (a + b)] - ab(a + b)

= bc(b + c) + ac(b + c) - ac(a + b) - ab(a + b)

= c(b + c)(a + b) - a(a + b)(b + c)

= (a + b)(b + c)(c - a)