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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

= x²y+xy²+y²z+yz²+x²z+xz²+2xyz

=(x²y+x²z+xz²+xyz) + ( xy²+y²z+yz²+xyz)

=x(xy+xz+z²+yz)+y(xy+yz+z²+xz)

=(xy+xz+yz+z²).(x+y)

=(x(y+z)+z(y+z)).(x+y)

=((y+z).(x+z)).(x+y)= (x+y)(x+z)(y+z)

2. 3(x-3)(x-7)+(x-4)²+48

=3(x²+4x-21)+x²-8x+16+48

=4x²-4x+1 = (2x-1)²

Thay x=0,5 vào bt trên, ta có : (2.0,5 -1)²=0

3, x²-6x+10

= x²-2.3.x+9+1

=(x-3)²+1 \(\ge\)1 >0 ( do (x-3)² >=0 với mọi x)

=> x²6x+10 >0 với mọi x

4x-x²-5

=-(x²-4x+5)

=- (x²-2.2x+4+1)

= - ((x-2)²+1) = -(x-2)²-1\(\le\)-1 < 0 ( do (x-2)²\(\ge\)0 với mọi x => - (x-2)²\(\le\)0 với mọi x)

vậy, 4x-x²-5<0 với mọi x

Ta có : x² - 6x + 10 

= x² - 6x + 9 + 1 

= (x - 3)² + 1

Mà (x - 3)² \(\ge0\forall x\)

Nên : (x - 3)² + 1 \(\ge1\forall x\)

=> (x - 3)² + 1 \(>0\)(đpcm)

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)