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Ta có :
A=xy(x+y+z) -xyz +(yz+zx)(x+y+z)
A=xy(x+y+z -z) + z(x+y)(x+y+z)
A=xy(x+y) +z(x+y)(x+y+z)
A=(x+y)(xy+z(x+y+z))
A=(x+y)(xy+zx+z(y+z))
A=(x+y)(x(y+z)+z(y+z))
A=(x+y)(y+z)(x+z)
P/ s : ko biết có đúng ko
Ta có
A=(x+y)(xy+yz+xz)+z(xy+yz+xz)-xyz
=(x+y)(xy+yz+xz)+xyz+z^2(y+x)-xyz
= (x+y)(xy+yz+xz + z^2)
= (x+y)(y+z)(x+z)
k cho mk đi
(xy+yz+zx)(x+y+z)-xyz
= (xy+yz)(x+y+z)+ x2z+xyz+ xz2-xyz
= (xy+yz)(x+y+z)+ x2z + xz2
= (x+z)(xy+y2+zy)+ xz(x+z) = (x+z)(xy+y2+zy+xz) = (x+z)(x+y)(x+z)
\(xy+yz+zx-xyz=1-x-y-z+xy+yz+zx-xyz\)
\(=\left(1-x\right)-y\left(1-x\right)-z\left(1-x\right)+yz\left(1-x\right)\)
\(=\left(1-x\right)\left(1-y-z+yz\right)=\left(1-x\right)\left(1-y\right)\left(1-z\right)\)
\(xy+yz+zx+xyz+2=1+x+y+z+xy+yz+zx+xyz\)
\(=\left(1+x\right)+y\left(1+x\right)+z\left(1+x\right)+yz\left(1+x\right)\)
\(=\left(1+x\right)\left(1+y\right)\left(1+z\right)\)
\(1+x+y+z=1+1\Rightarrow1+x=\left(1-y\right)+\left(1-z\right)\ge2\sqrt{\left(1-y\right)\left(1-z\right)}\)
Tương tự ta cũng có: \(1+y\ge2\sqrt{\left(1-z\right)\left(1-x\right)}\)
\(1+z\ge2\sqrt{\left(1-x\right)\left(1-y\right)}\)
Vậy \(S\le\frac{\left(1-x\right)\left(1-y\right)\left(1-z\right)}{8\left(1-x\right)\left(1-y\right)\left(1-z\right)}=\frac{1}{8}\)
xy(x+y+z) -xyz +(yz+zx)(x+y+z)
=xy(x+y+z -z) + z(x+y)(x+y+z)
=xy(x+y) +z(x+y)(x+y+z)
=(x+y)(xy+z(x+y+z))
=(x+y)(xy+zx+z(y+z))
=(x+y)(x(y+z)+z(y+z))
=(x+y)(y+z)(x+z)