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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`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
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
ta có A+B+C=x2yz+xy2z+xyz2
=x(xyz)+y(xyz)+z(xyz)
=x.1+y.1+z.1
=x+y+z(dpcm)
\(A=x^2yz=x.\left(xyz\right)=x.1=x\)
\(B=xy^2z=y.\left(xyz\right)=y.1=y\)
\(C=xyz^2=z.\left(xyz\right)=z.1=z\)
\(\Rightarrow A+B+C=x+y+z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A+B+C=\(X^2\)YZ+X\(Y^2\)Z+XY\(Z^2\)=XXYZ+XYYZ+XYZZ=(X+Y+Z)XYZ
MÀ XYZ=1=>A+B+C=(X+Y+Z)*1=X+Y+Z
![](https://rs.olm.vn/images/avt/0.png?1311)
A=x^2yz
B=xy^2z
C=xyz^2
=>A+B+C=x^2yz+xy^2z+xyz^2=xyz(x+y+z)=xyz
\(A+B+C=xyz\)
\(VT=A+B+C\)
\(\Leftrightarrow VT=x^2yz+xy^2z+xyz^2\)
\(\Leftrightarrow VT=xyz\left(x+y+z\right)\)
\(\Leftrightarrow VT=xyz\)
\(\Rightarrow VT=VP\)
\(\Rightarrow A+B+C=xyz\left(dpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A + B + C = x2.y.z + x.y2.z + x.y.z2 = x.y.z.(x + y + z) = x.y.z .1 = xyz (Vì x+ y + z = 1)
Ta có:
\(A+B+C=x^2yz+xy^2z+xyz^2\\ A+B+C=xyz\left(x+y+z\right)\\ A+B+C=xyz\times1\\ A+B+C=xyz\)
Vậy A+B+C=xyz