`A + B + C = x^2yz + xy^2z + zy^2x = xyz(x+y+z) = xyz`.

\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz\)

Vậy ta có đpcm 

lỗi r

Ta có : 

\(A+B+C\)

\(=x^2yz+xy^2z+xyz^2\)

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

\(=xyz.1\)

\(=xyz\left(đpcm\right)\)

ta có: A + B + C = x²yz +xy²z + xyz² = xyz.(x+y+z) = xyz.1=xyz

=> A+B+C = xyz

#

Bài 4: 

b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)

c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)

Câu 1:

\(A\left(x\right)+B\left(x\right)\)

\(=\left(6x-4x^3+x-1\right)+\left(-3x-2x^3-5x^2+x+2\right)\)

\(=\left(6x+-3x+x\right)-\left(4x^3+2x^3\right)-5x^2+\left(-1+2\right)\)

\(=-6x^3-5x^2+4x+1\)

\(A\left(x\right)-B\left(x\right)\)

\(=\left(6x-4x^3+x-1\right)-\left(-3x-2x^3-5x^2+x+2\right)\)

\(=\left(-4x^3+2x^3\right)+5x^2+\left(6x+x-x\right)+\left(-1-2\right)\)

\(=-2x^3+5x^2+6x-3\)

\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz\)

\(A=x^2yz\) \(B=xy^2z\) \(C=xyz^2\)

\(A+B+C=x^2yz+xy^2z+xyz^2\)

                    \(=xyz\left(x+y+z\right)=xyz.1=xyz\)

\(Ta\) \(có:\)

\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz.1=xyz\)

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lỡ tay ấn nhầm rùi 

a: A = -2xy + 3/2xy^2 + 1/2xy^2 + xy = -2xy + 2xy^2 + xy = 2xy^2 - xy

b: B = xy^2z + 2xy^2z - xyz - 3xy^2z + xy^2z = 3xy^2z - xyz

c: C = 4x^2y^3 + x^4 - 2x^2 + 6x^4 - x^2y^3 = 7x^4 + 3x^2y^3 - 2x^2

d: D = 3/4xy^2 - 2xy - 1/2xy^2 + 3xy = 5/4xy^2 + xy

e: E = 2x^2 - 3y^3 - z^4 - 4x^2 + 2y^3 + 3z^4 = -2x^2 - y^3 + 2z^4

f: F = 3xy^2z + xy^2z - xyz + 2xy^2z - 3xyz = 6xy^2z - 2xyz

a: A=-2xy+3/2xy^2+1/2xy^2+xy

=-2xy+xy+3/2xy^2+1/2xy^2

=2xy^2-xy

b: \(B=xy^2z+2xy^2z-xyz-3xy^2z+xy^2z\)

\(=xy^2z\left(1+2-3+1\right)-xyz=xy^2z-xyz\)

c: \(=4x^2y^3-x^2y^3+x^4+6x^4-2x^2\)

\(=7x^4-x^2+3x^2y^3\)

d: \(=\dfrac{3}{4}xy^2-\dfrac{1}{2}xy^2+3xy-2xy\)

=1/4xy^2+xy

e: \(=2x^2-4x^2-3y^3+2y^3+3z^4-z^4\)

\(=-2x^2-y^3+2z^4\)

f: \(=xy^2z+3xy^2z+2xy^2z-xyz-3xyz\)

\(=6xy^2z-4xyz\)