Phân tích đa thức thành nhân tử:
x2y2 ( y -x ) + y2z2( z - y ) - z2x2( z -x )
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\(3,=\left(x-y\right)^3+\left(y-x+x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3+\left(y-x\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-x+x-z\right)+\left(x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3-\left(x-y\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-z\right)-\left(z-x\right)^3+\left(z-x\right)^3\\ =3\left(y-x\right)\left(x-z\right)\left(y-z\right)\)
\(4,=\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-3x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)
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
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)-3xzy+z^3\)
\(=\left(-z\right)^3-3xy\cdot\left(-z\right)-3xyz+z^3=z^3+3xyz-3xyz-z^3=0\)
yz(y+z)+zx(z+x)+xy(x+y)+2xyz
=yz(y+z+x-x)+zx(z+x+x-x)+xy(x+y)+2xyz
=(x-z)(yz+zx)+(x+y)(yz+xy)+2x2z+2xyz
=y(x-z)(x+y)+y(x+z)(x+y)+2xz(x+y)
=(x+y)[y(x-z)+y(x+z)+2xz]
=(x+y)(yx-yz+yx+yz+2xz)
=(x+y)2x(y+z)
\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2\left(z-y+y-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2\left(z-y\right)+y^2\left(y-x\right)+z^2\left(x-y\right)\)
\(=\left[x^2\left(y-z\right)-y^2\left(y-z\right)\right]+\left[y^2\left(y-x\right)-z^2\left(y-x\right)\right]\)
\(=\left(x^2-y^2\right)\left(y-z\right)+\left(y^2-z^2\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(y-z\right)+\left(y-z\right)\left(y+z\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left[\left(x+y\right)-\left(y+z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
Ta có: M = 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)(x + y)
= (x + y)(xy + zx + zy + z2)
= (x + y)[x(y + z) + z(y + z)]
M = (x + y)(y + z)(z + x) (đpcm)
Ta có:
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-x-y+x\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-x\right)-y^2z^2\left(y-x\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(y-x\right)\left(x-z\right)\left(x+z\right)+z^2\left(z-x\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y^2x+y^2z-z^2y-z^2x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y-z\right)\left(xy+yz+zx\right)\)