Xét tích : \(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)

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

\(+z^3\left(y-x\right)+z^2\left(y-x\right)\left(y+x\right)\)

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

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

\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)

Như vậy:

 \(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)

<=> \(\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)

Ta có: \(\frac{\frac{x^2\left(z-y\right)}{yz}+\frac{y^2\left(x-z\right)}{xz}+\frac{z^2\left(y-x\right)}{xy}}{\frac{x\left(z-y\right)}{yz}+\frac{y\left(x-z\right)}{xz}+\frac{z\left(y-x\right)}{xy}}\)

 \(=\frac{\frac{x^3\left(z-y\right)}{xyz}+\frac{y^3\left(x-z\right)}{xyz}+\frac{z^3\left(y-x\right)}{xyz}}{\frac{x^2\left(z-y\right)}{xyz}+\frac{y^2\left(x-z\right)}{xyz}+\frac{z^2\left(y-x\right)}{xyz}}\)

\(=\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)

Bạn xem lại đề nhé :)

Thay 1 bằng xy + yz + zx được : 

\(1+y^2=xy+yz+zx+y^2=x\left(y+z\right)+y\left(y+z\right)=\left(x+y\right)\left(y+z\right)\)

Tương tự : \(1+x^2=\left(x+y\right)\left(x+z\right)\), \(1+z^2=\left(x+z\right)\left(z+y\right)\)

Suy ra \(Q=x\sqrt{\frac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+y\right)\left(x+z\right).\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\frac{\left(x+y\right)\left(x+z\right).\left(x+y\right)\left(y+z\right)}{\left(x+z\right)\left(z+y\right)}}\)

\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}=x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)

\(=2\left(xy+yz+zx\right)=2\)(vì x,y,z > 0)

Giúp mình với các bạn

\(1A=\frac{xy}{\left(z-x\right)\left(z-y\right)}+\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{zx}{\left(y-x\right)\left(y-z\right)}\)

\(=-1\left(\frac{xy}{\left(y-z\right)\left(z-x\right)}+\frac{yz}{\left(x-y\right)\left(z-x\right)}+\frac{zx}{\left(y-z\right)\left(x-y\right)}\right)\)

\(=-1.\left(\frac{xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\right)\)

 \(=\frac{-1\left(x-y\right)\left(z-x\right)\left(z-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)

\(A=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)

\(=\frac{z\left(y^2-x^2\right)+z^2\left(x-y\right)+xy\left(x-y\right)}{ }=\frac{\left(x-y\right)z\left(z-x-y\right)+xy\left(x-y\right)}{ }=\frac{\left(x-y\right)\left(z^2-xz-yz+xy\right)}{ }=\)

\(=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{ }=1\)