Question

phân tích thành nhân tử:

X=ab(a+b)+bc(b+c)+ca(c+a)+2abc

Answer 1

X=ab(a+b)+bc(b+c)+ca(c+a)+2abc

=a²b+ab²+b²c+bc²+c²a+a²c+2abc

=b(a+c)²+b²(a+c)+ac(a+c)+2abc

=[b(a+c)+b²+ac](a+c)+2abc

=(ab+bc+b²+ac)(a+c)+2abc

=[b(c+b)+a(b+c)](a+c)+2abc

=[(b+a)(b+c)](a+c)+2abc

rồi j nữa, tịt rồi

Answer 2

giúp đi

Answer 3

X=ab(a+b)+bc(b+c)+ca(c+a)+2abc

=a²b+ab²+b²c+bc²+ac²+a²c+2abc

=a(ab+b²+c²+ac+2bc)+bc(b+c)

=a[(b+c)²+a(b+c)]+bc(b+c)

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

=a(b+c)(a+b+c+bc)
Xin hết

Answer 4

X=ab(a+b)+bc(b+c)+ca(c+a)+2abc
  = a²b+ab²+b²c+bc²+ac²+a²c+2abc
  = a(ab+b²+c²+ac+2bc)+bc(b+c)
  =a[a(b+c)+(b+c)² ] +bc(b+c)
  =a(a+b+c)(b+c)+ bc(b+c)
  =(b+c) (a²+ab+ac+bc)
  =(b+c) (a+b)(b+c)