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

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

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

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

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

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

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

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

Vậy A = 1

\(\left(x-1\right)^2\ge0\Rightarrow x^2-2x+1\ge0\Rightarrow x^2+1\ge2x\)

\(\left(y-2\right)^2\ge0\Rightarrow y^2-4y+4\ge0\Rightarrow y^2+4\ge4y\)

\(\left(z-3\right)^2\ge0\Rightarrow z^2-6z+9\ge0\Rightarrow z^2+9\ge6z\)

Do đó: \(\left(x^2+1\right)\left(y^2+4\right)\left(z^2+9\right)\ge2x.4y.6z=48xyz\)

Dấu "=" xảy ra khI: \(\hept{\begin{cases}x-1=0\\y-2=0\\z-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}}\)

Vậy \(C=\frac{1^3+2^3+3^3}{\left(1+2+3\right)^3}=\frac{6^2}{6^3}=\frac{1}{6}\)

Chúc bạn học tốt.

Đăt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\Rightarrow x=2k,y=3k,z=4k\)

\(\Rightarrow M=\frac{y+x-z}{x-y+z}=\frac{3k+2k-4k}{2k-3k+4k}=\frac{k}{3k}=\frac{1}{3}\)

Thank you!!!!

1) 

Nếu x>1 thì x^2>1; y^2;z^2 cx lớn=1

=> x^2+y^2+z^2>1=> Loại

Nếu x<-1=> x^2>1; y^2;z^2 cx lớn=1

=> x^2+y^2+z^2>1=> Loại

CMTT vs y,z thì -1<=x,y,z<=1

TH1: -1<=x<0

=> x<x^2 do x âm và x^2 dương

CMTT => y<y^2; z<z^2

=> x+y+z<x^2+y^2+z^2

Mà x+y+z=1, x²+y²+z²=1=> x+y+z=x^2+y^2+z^2

=> LOẠI.

TH2: 0<=x,y,z<=1

=> x>=x^2; y>=y^2; z>=z^2

=> x+y+z>=x^2+y^2+z^2

Mà x+y+z=1, x²+y²+z²=1=> x+y+z=x^2+y^2+z^2

=> ''='' xảy ra <=> x=0 hoặc 1; y=0 hoặc 1; z=0 hoặc 1

=> (x,y,z)=(0;0;1) và các hoán vị

=> A=1.

Khó quá đúng ko