\(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)

\(=x^4y-x^4z+y^4z-y^4x+z^4\left(x-y\right)\)

\(=xy\left(x^3-y^3\right)-z\left(x^4-y^4\right)+z^4\left(x-y\right)\)

\(=xy\left(x-y\right)\left(x^2+xy+y^2\right)-z\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)+z^4\left(x-y\right)\)

\(=\left(x-y\right)\left[xy\left(x^2+xy+y^2\right)-z\left(x^3+x^2y+xy^2+y^3\right)+z^4\right]\)

\(=\left(x-y\right)\left(x^3y+x^2y^2+xy^3-x^3z-x^2yz-xy^2z-y^3z+z^4\right)\)

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

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

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

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

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

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

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

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

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

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

\(Ta\)\(có\)\(x>y>z\Rightarrow\left(x-y\right);\left(y-z\right);\left(x-z\right)>0\)

                 \(\left(x+z\right)^2;\left(y+z\right)^2;\left(x+y\right)^2\ge0\)

\(\Rightarrow A>o\Rightarrow A\)\(luôn\)\(dương\)

\(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)

\(A=x^4\left(y-z\right)+y^4\left(z-x\right)-z^4\left[\left(y-z\right)+\left(z-x\right)\right]\)

\(A=x^4\left(y-z\right)-z^4\left(y-z\right)+y^4\left(z-x\right)-z^4\left(z-x\right)\)

\(A=\left(y-z\right)\left(x^4-z^4\right)+\left(z-x\right)\left(y^4-z^4\right)\)

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

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

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

\(A=\frac{1}{2}\left(x-y\right)\left(y-z\right)\left(x-z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\right]\)

Vì \(x>y>z\Rightarrow A>0\)

1 c nha các bạn

Ta có:\(P=x^3\left(z-y^2\right)+y^3x-y^3z^2+z^3y-z^3x^2+x^2y^2z^2-xyz\)

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

\(\Rightarrow P=x^3\left(z-y^2\right)+x^2z^2\left(y^2-z\right)-yz^2\left(y^2-z\right)+xy\left(y^2-z\right)\)

\(\Rightarrow P=\left(y^2-z\right)\left(x^2z^2-x^3-yz^2+xy\right)\)

\(\Rightarrow P=\left(y^2-z\right)\left(x^2z^2-x^3+xy-yz^2\right)\)

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

\(\Rightarrow P=\left(y^2-z\right)\left(x^2\left(z^2-x\right)-y\left(z^2-x\right)\right)\)

\(\Rightarrow P=\left(y^2-z\right)\left(z^2-x\right)\left(x^2-y\right)\)

\(\Rightarrow P=abc\)

Vì a, b, c là hằng số nên P có giá trị không phụ thuộc vào x, y, z

câu nào cx ghi là lớp 8 nhưng thực ra lớp 9 cx k nổi vc

lớp 8 đó anh Thắng ạ =.="

Sửa đề z^4(z-y) thành z^4(x-y)

Đặt \(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)

\(=x^4\left(y-x+x-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)

\(=-x^4\left(x-y\right)+x^4\left(x-z\right)-y^4\left(x-z\right)+z^4\left(x-y\right)\)

\(=\left(x-y\right)\left(z^4-x^4\right)+\left(x-z\right)\left(x^4-y^4\right)\)

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

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

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

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

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

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

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

\(=\frac{1}{2}\left(x-y\right)\left(x-z\right)\left(y-z\right)\left(2x^2+2y^2+2z^2+2xy+2yz+2xz\right)\)

\(=\frac{1}{2}\left(x-y\right)\left(x-z\right)\left(y-z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\right]\)

Vì \(x>y>z\Rightarrow\hept{\begin{cases}x-y>0\\x-z>0\\y-z>0\end{cases}}\) và \(\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\ge0\)

=>....

mình nghĩ ra cách này ko biết đúng hay sai, nhưng mình sẽ cm cho bạn xem trước cái này để mình đảo lại trong quá trình làm bài luôn cho đỡ mất thời gian

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

thế nên sẽ đảo ngược lại trong bài này, vây ta sẽ có

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

thay vào đề bài ta được

\(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(y-x\right)}\\ =\dfrac{1}{x-y}-\dfrac{1}{x-z}+\dfrac{1}{y-z}-\dfrac{1}{y-x}+\dfrac{1}{z-x}-\dfrac{1}{y-x}\\ =\dfrac{1}{x-y}+\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{y-z}+\dfrac{1}{z-x}+\dfrac{1}{z-x}\\ =\dfrac{2}{x-y}+\dfrac{2}{y-x}+\dfrac{2}{z-x}\left(đpcm\right)\)

vậy ...

mình nghĩ ra thì là như z, chúc may mắn :)

bài này mk cũng làm dc ròi

thanks bạn nha