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

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

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

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

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

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

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

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

Vì \(x>y>z\Rightarrow A>0\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a: Ta có: \(\left(ac+bd\right)^2-\left(ad+bc\right)^2\)

\(=a^2c^2+b^2d^2+2abcd-a^2d^2-b^2c^2-2abcd\)

\(=a^2\left(c^2-d^2\right)-b^2\left(c^2-d^2\right)\)

\(=\left(a^2-b^2\right)\left(c^2-d^2\right)\)

Bạn có làm đc câu b ko, nếu đc thì làm nốt giùm mink nha

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 =.="

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