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vì x - y - z = 0 nên x = y + z

Xét tổng A + B = xyz - xy² - xz² + y³ + z³

= ( y + z ) . yz - ( y + z ) . y² - ( y + z ) . z² + y³ + z³

= y²z + yz² - y³ - y²z - yz² - z³ + y³ + z³ = 0

Vậy ...

x-y-z=0

=>x=y+z

=>x²=y²+z²+2yz

=>y²+z²=x²-2yz

*A=xyz-xy²-xz²=x.(yz-y²-z²)=x.[yz-(x²-2yz)]=x.(3yz-x²)=3xyz-x³

*B=y³+z³=(y+z)(x²-yz+z²)=x.(x²-2yz-yz)=x³-3xyz=-(3xyz-x³)

Vậy A và B đối nhau

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

\(B=y^3+z^3=\left(y+z\right)\left(y^2-yz+z^2\right)\)

Lại có \(x-y-z=0\)\(\Leftrightarrow\)\(y+z=x\)

\(\Rightarrow\)\(B=\left(y+z\right)\left(y^2-yz+z^2\right)=x\left(y^2-yz+z^2\right)\) là số đối của \(A\) ( đpcm ) 

Chúc bạn học tốt ~ 

Vì x-y-z=0 nên x=y+z

Xét tổng A+B=xyz-xy²-xz²+y³+z³

= (y+z).yz-(y+z).y²-(y+z).z²+y³+z³

=y².z+y.z²-y³-y².z-yz²-z³+y³+z³

=(yz²-yz²)+(y³-y³)+(y²z-y²z)+(z³-z³)

=0+0+0+0=0

Vay A và B la hai da thuc doi nhau

Giải:

Ta có: \(B=y^3+z^3\)

\(=y^2.\left(x-z\right)+z^2.\left(x-y\right)\)

\(=xy^2-y^2z+xz^2-yz^2\)

\(=xy^2+xz^2-\left(y^2z+yz^2\right)\)

\(=xy^2+xz^2-yz.\left(y+z\right)\)

\(=xy^2+xz^2-xyz\)

\(=-A\)

Do đó: \(A\) và \(B\) là 2 đa thức đối nhau.