Question

Phân tích thành nhân tử

xy(x+y)+yz(y+z)+xz(x+z)+2xyz

Answer 1

xy(x+y)+yz(y+z)+xz(x+z)+2xyz 
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 
= xy(x + y) + yz(y + z + x) + xz(x + z + y) 
= xy(x + y) + z(x + y + z)(y + x) 
= (x + y)(xy + zx + zy + z²) 
= (x + y)[x(y + z) + z(y + z)] 
= (x + y)(y + z)(z + x)

Answer 2

Ta co : 

Đặt tổng trên là A 

A= xy(x+y)+yz(y+z)+xz(x+z)+2xyz  

A= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz  

A= xy(x + y) + yz(y + z + x) + xz(x + z + y)  

A= xy(x + y) + z(x + y + z)(y + x)  

A= (x + y)(xy + zx + zy + z² )   

A= (x + y)[x(y + z) + z(y + z)]  

A= (x + y)(y + z)(z + x)