Bài 3: 

\(B=x^4-4x^3-2x^2+12x+9\)

\(=x^4-3x^3-x^3+3x^2-5x^2+15x-3x+9\)

\(=\left(x-3\right)\left(x^3-x^2-5x-3\right)\)

\(=\left(x-3\right)\left(x^3-3x^2+2x^2-6x+x-3\right)\)

\(=\left(x-3\right)^2\cdot\left(x+1\right)^2\)

\(=\left(x^2-2x-3\right)^2\)

Bài 3: 

\(B=x^4-4x^3-2x^2+12x+9=\left(x^4+x^3\right)-\left(5x^3+5x^2\right)+\left(3x^2+3x\right)+\left(9x+9\right)=\left(x^3-5x^2+3x+9\right)\left(x+1\right)=\left[\left(x^3+x^2\right)-\left(6x^2+6x\right)+\left(9x+9\right)\right]\left(x+1\right)=\left(x^2-6x+9\right)\left(x+1\right)^2=\left(x-3\right)^2\left(x+1\right)^2=\left[\left(x-3\right)\left(x+1\right)\right]^2\)

\(C=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2\)

\(=4x\left(x+y+z\right)\left(x+y\right)\left(x+z\right)+y^2z^2\)

\(=4\left(x^2+xy+xz\right)\left(x^2+xy+xz+yz\right)+y^2z^2\left(1\right)\)

Đặt \(a=x^2+xy+xz\)và \(b=yz\)ta có:

\(\left(1\right)\Rightarrow C=4a\left(a+b\right)+b^2=b^2+4ab+4a^2=\left(b+2a\right)^2\)

Vậy C là một số chính phương.

abcdacscas

B3 : t chỉ m r á :3
B4 : 
Ta có :
C= 4x ( x + y ) ( x + y + z ) ( y + z ) + y²x²
   = 4x ( x + y + z ) ( x + y ) ( x + z ) + y²x²
   = 4 ( x² + xy + xz ) ( x² + xy + xz + yz ) + y²x²
Đặt a = x² + xy + xz và b= yz , ta có :
  ⇒ C = 4a( a + b ) + b²
          = b² + 4ab + 4a²
          = ( b + a )²
  ⇒ C là số chính phương 
Chúc mừng m đã ghi xong bài , nhớ tick cho t nhoa bff!
            

\(C=4x\left(x+y+z\right)\left(x+y\right)\left(x+z\right)+y^2z^2=4\left(x^2+xy+xz\right)\left(x^2+xz+xy+yz\right)+y^2z^2\)

Đặt \(x^2+xy+xz=t\), ta có:

\(C=4t\left(t+yz\right)+y^2z^2=4t^2+4tyz+y^2z^2=\left(2t\right)^2+2.2t.yz+\left(yz\right)^2=\left(2t+yz\right)^2=\left(2x^2+2xy+2xz+yz\right)^2\)

=>đpcm

\(M=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2=4\left(x^2+xy+xz\right)\left(x^2+xy+xz+yz\right)+y^2z^2\)

Đặt \(x^2+xy+xz=a\) , ta có:

\(M=4a\left(a+yz\right)+y^2z^2=4a^2+4ayz+y^2z^2=\left(2a+yz\right)^2\)

\(M=\left(2x^2+2xy+2xz+yz\right)^2\)là số chính phương với \(x;y;z\in N\)