Bài 3: 

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

Bài 3: 

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

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

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

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

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

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

\(3,=\left(x-y\right)^3+\left(y-x+x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3+\left(y-x\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-x+x-z\right)+\left(x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3-\left(x-y\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-z\right)-\left(z-x\right)^3+\left(z-x\right)^3\\ =3\left(y-x\right)\left(x-z\right)\left(y-z\right)\)

\(4,=\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-3x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)

thay x=1 ;y=-1;z=2 vào biểu thức b) ta được:1.-1+(-1)\(^{^2}\).2\(^2\)+\(2^3\).\(1^3\)

=-1+1.4+8.1

=-1+4+8=11

Thay x = 1 y = -1 z = 2 ta có

1.(-1) + (-1)².2² + 2³ .1³ = 1.(-1) + 1 .4 + 8 .1 = 11

a: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

=>\(\frac{yz+xz+xy}{xyz}=0\)

=>xy+xz+yz=0

Đặt a=xy; b=xz; c=yz

=>a+b+c=0

=>\(\left(a+b+c\right)^2=0\)

=>\(a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left\lbrack a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\right\rbrack\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left(a^2b^2+b^2c^2+a^2c^2\right)\)

Ta có: \(\left(a^2+b^2+c^2\right)=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)

=>\(2\left(a^2b^2+b^2c^2+a^2c^2\right)=\left(a^2+b^2+c^2\right)^2-\left(a^4+b^4+c^4\right)\)

=>\(4\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)\)

DO đó, ta có: \(\left(a^2+b^2+c^2\right)^2=2\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)\)

=>\(2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

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

b:

x+y+z=0

=>x+y=-z

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)-3xzy+z^3\)

\(=\left(-z\right)^3-3xy\cdot\left(-z\right)-3xyz+z^3=z^3+3xyz-3xyz-z^3=0\)

bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc

1). x²y²(y-x)+y²z²(z-y)-z²x²(z-x)

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

3)yz(y+z)+xz(z-x)-xy(x+y)

5)y(x-2z)²+8xyz+x(y-2z)²-2z(x+y)²

6)8x³(y+z)-y³(z+2x)-z³(2x-y)

7) (x2+y2)³+(z2-x2)³-(y2+z2)³

3 3 3 3 3 3 3 3 3 3 3 3 3