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

Tương tự: \(\dfrac{1}{\left(y-z\right)\left(x^2+xz-y^2-yz\right)}=\dfrac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}\)

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

\(\Rightarrow M=\dfrac{y-z-z+x-x+y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}\\ M=\dfrac{2}{\left(x-y\right)\left(z-x\right)\left(x+y+z\right)}\)

tại sao lại không có điều kiện ? 

(x+y+z)(xy+yz+zx)=xyz

x²y+xyz+zx²+xy²+y²z+xyz+xyz+yz²+z²x=xyz

(x²y+xy²)+(xyz+zx²)+(y²z+xyz)+(yz²+z²x)+xyz=xyz

xy(x+y)+zx(y+x)+yz(y+x)+z²(y+x)+xyz=xyz

(x+y)(xy+xz+yz+z²)+xyz=xyz

(x+y)[(xy+xz)+(yz+z²)]+xyz=xyz

(x+y)[x(y+z)+z(y+z)]+xyz=xyz

(x+y)(x+z)(y+z)+xyz=xyz

(x+y)(x+z)(y+z)=xyz-xyz

(x+y)(x+z)(y+z)=0

=>\(\left[{}\begin{matrix}x+y=0\\x+z=0\\y+z=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-y\\x=-z\\y=-z\end{matrix}\right.\)

Với x=-z

=>VT= x²⁰¹⁵+y²⁰¹⁵+z²⁰¹⁵=(-z)²⁰¹⁵+z²⁰¹⁵+y²⁰¹⁵=y²⁰¹⁵

VP=(x+y+z)²⁰¹⁵=(-z+y+z)²⁰¹⁵=y²⁰¹⁵

Vậy x²⁰¹⁵+y²⁰¹⁵+z²⁰¹⁵=(x+y+z)²⁰¹⁵ với (x+y+z)(xy+yz+zx)=xyz

Tôi không biết làm . Khó quá !!!!!!!!!!!!!!

các bạn cố gắng giúp tôi đi mà . tôi cần gấp lắm !!!!!!!!!!!!!!!!!!!