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

\(A=x^2yz\) \(B=xy^2z\) \(C=xyz^2\)

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

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

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Ta có:

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

\(B=xy^2z=x.y.y.z=y.xyz\left(2\right)\)

\(C=xyz^2=x.y.z.z=z.xyz\left(3\right)\)

Lấy (1)+(2)+(3),vế theo vế ta được:

\(A+B+C=x.xyz+y.xyz+z.xyz=\left(x+y+z\right).xyz=xyz\) (vì x+y+z=1)

Vậy A+B+C=xyz      (đpcm)

ta có A+B+C=x²yz+xy²z+xyz²

^{=x(xyz)+y(xyz)+z(xyz)}

^=x.1+y.1+z.1

^=x+y+z(dpcm)

\(A=x^2yz=x.\left(xyz\right)=x.1=x\)

\(B=xy^2z=y.\left(xyz\right)=y.1=y\)

\(C=xyz^2=z.\left(xyz\right)=z.1=z\)

\(\Rightarrow A+B+C=x+y+z\)

Ta có \(A+B+C=x^2yz+xy^2z+xyz^2\)

\(A+B+C=\left(xyz\right)x+\left(xyz\right)y+\left(xyz\right)z\)

\(A+B+C=\left(x+y+z\right)xyz\)

Mà xyz=1 thay vào A+B+C ta có 

\(A+B+C=x+y+z\) (đpcm)

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

Ta có : \(A+B+C=x^2yz+xy^2z+xyz^2\)

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

\(=xyz\left(đpcm\right)\)

admin là ai

A + B + C = x²yz + xy²z + xyz²

suy ra : xyz ( x + y + z ) 

= xyz

Theo đầu bài ta có:
\(\hept{\begin{cases}A=x^2yz=xyz\cdot x\\B=xy^2z=xyz\cdot y\\C=xyz^2=xyz\cdot z\end{cases}}\)
\(\Rightarrow A+B+C=xyz\cdot x+xyz\cdot y+xyz\cdot z\)
\(\Rightarrow A+B+C=xyz\left(x+y+z\right)\)
Mà \(x+y+z=1\Rightarrow A+B+C=xyz\) ( đpcm )

Ta có 

\(\hept{\begin{cases}A=x^2yz=xyz.x\\B=xy^2z=xyz.y\\C=xyz^2=xyz.z\end{cases}}\)

\(\Rightarrow A+B+C=xyz.x+xyz.y+xyz.z\)

\(\Rightarrow A+B+C=xyz.\left(x+y+z\right)\)

Mà \(x+y+z=1\Rightarrow A+B+C=xyz\)

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

\(A+B+C=xyz\)

\(VT=A+B+C\)

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

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

\(\Leftrightarrow VT=xyz\)

\(\Rightarrow VT=VP\)

\(\Rightarrow A+B+C=xyz\left(dpcm\right)\)