\(\left(x+y+z\right)\left(xy+yz+xz\right)-xyz=xy\left(x+y+z\right)-xyz+\left(yz+xz\right)\left(x+y+z\right)\)

\(=xy\left(x+y+z-z\right)+z\left(x+y\right)\left(x+y+z\right)\)

\(=xy\left(x+y\right)+z\left(x+y\right)\left(x+y+z\right)\)

\(=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)

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

\(=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)

\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

Ta có : \(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)

\(=\left[xy\left(x+y\right)+xyz\right]+\left[yz\left(y+z\right)+xyz\right]+xz\left(x+z\right)\)

\(=xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)\)

\(=y\left(x+y+z\right)\left(x+z\right)+xz\left(x+z\right)\)

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

\(=\left(x+z\right)\left(x+y\right)\left(y+z\right)\)

\(xy\left(x-y\right)-xz\left(x+z\right)+yz\left(2x-y+z\right)\)

\(=xy\left(x-y\right)-xz\left(x+z\right)+yz\left(x-y+x+z\right)\)

\(=xy\left(x-y\right)-xz\left(x+z\right)+yz\left(x-y\right)+yz\left(x+z\right)\)

\(=\left(x-y\right)\left(xy+yz\right)-\left(xz-yz\right)\left(x+z\right)\)

\(=y\left(x-y\right)\left(x+z\right)-z\left(x-y\right)\left(x+z\right)\)

\(=\left(y-z\right)\left(x-y\right)\left(x+z\right)\)

Ta có :

xy(x−y)−xz(x+z)+yz(2x−y+z)

=xy(x−y)−xz(x+z)+yz(x−y+x+z)

=xy(x−y)−xz(x+z)+yz(x−y)+yz(x+z)

=(x−y)(xy+yz)−(xz−yz)(x+z)

=y(x−y)(x+z)−z(x−y)(x+z)

=(y−z)(x−y)(x+z)

P/s tham khảo nha

\(yz\left(y+z\right)+xz\left(z-x\right)-xy\left(x+y\right)\)

\(=-[xy(x+y)-yz(y+z)-zx(z-x)]\)

\(=-(y.[x(x+y)-z(y+z)]-zx(z-x))\)

\(=-[y.(x^2+xy-zy-z^2)-zx(z-x)]\)

\(=-[y.(x^2-z^2+xy-zy)-zx(z-x)]\)

\(=-(y.[(x+z)(x-z)+y.(x-z)]-zx(z-x))\)

\(=-[y.(x-z)(x+z+y)+zx(x-z)]\)

\(=[(x-z)[y(x+z+y)+zx]]\)

\(=-(x-z)(yx+yz+y2+zx)\)

\(=-(x-z)(yx+zx+yz+y2)\)

\(=-[(x-z)[x.(y+z)+y.(y+z)]]\)

\(=-(x-z)(y+z)(x+y)\)

=2x(y-z) - (y-z)(x+t)

=(2x -x +t).(y-z)

Goi da thuc tren la A

Thay a=b -> A= 0 -> A chua nghiem la a-b

Tuong tu b=c-> A = 0 - > A chua nghiem la b -c

Tuong tu c =a - > A = 0 -> A chua nghiem la c-a

=> A = k(a - b)(b - c)(c - a)

Vì A có bậc 3 mà (a - b)(b - c)(c - a) cũng có bậc 3 -> k là 1 số 

Thay a = 3, b= 2, c= 1

=> A= -6=k.1.1..-2

=> k = 3

=> A = 3(a - b)(b - c)(c - a)

Đây gọi là phương pháp giá trị riêng bạn nha!

x^5 + x + 1

= x^5 - x^2 + (x^2 + x + 1)

= x^2(x^3 - 1) + ( x^2 + x + 1)

= x^2( x - 1)(x^2 + x + 1) + ( x^2 + x + 1)

= (x^3 - x^2 + 1)(x^ 2 + x + 1)