Ta có : \(x+y+z=0\Rightarrow x+y=-z\)

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

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

Tương tự ta có : \(y^2+z^2=x^2-2yz\)

\(x^2+z^2=y^2-2xz\)

Thay vào biểu thức ta có :

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

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

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

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

\(=\frac{3xyz}{2xyz}=-\frac{3}{2}\)

Chỗ \(x^3+y^3+z^3=3xyz\)là do \(x+y+z=0\)nhé, bạn cần chứng minh không ?

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

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

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

Lại vì \(x+y+z=0\Rightarrow\hept{\begin{cases}z-x=-2x-y\\x-y=-2y-z\\y-z=-x-2z\end{cases}\left(3\right)}\)

Thay (2) và (3) vào (1) ta được:

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

\(=\frac{x^2}{2y\left(x+y\right)}+\frac{y^2}{2z\left(y+z\right)}+\frac{z^2}{2x\left(x+z\right)}\left(4\right)\)

Thay (2) vào (4) ta được:
\(A=\frac{x^2}{-2yz}+\frac{y^2}{-2zx}+\frac{z^2}{-2xy}\)

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

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

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

\(=\frac{-3xyz}{-2xyz}=\frac{3}{2}\)

Vậy ...

X+y=z=0 chứ ko phải x+y=z

\(x+y+z=0\Rightarrow x+y=-z\)

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

Tương tự: \(y^2+z^2-x^2=-2yz,x^2+z^2-y^2=-2xz\)

\(\frac{1}{y^2+z^2-x^2}+\frac{1}{x^2+y^2-z^2}+\frac{1}{x^2+z^2-y^2}\)

\(=\frac{1}{-2yz}+\frac{1}{-2xy}+\frac{1}{-2xz}=\frac{x+y+z}{-2xyz}=0\)

\(x+y+z=0\) => \(x+y=-z\) => \(\left(x+y\right)^2=z^2\)

=> \(x^2+2xy+y^2=z^2\)

=> \(z^2-x^2-y^2=2xy\)

Tương tự:

   \(x^2-y^2-z^2=2yz\)

   \(y^2-z^2-x^2=2zx\)

Thay vào tính M ta có:

  \(M=\frac{x^2}{2yz}+\frac{y^2}{2zx}+\frac{z^2}{2xy}\)

        \(=\frac{1}{2}\left(\frac{x^3+y^3+z^3}{xyz}\right)\)     (*)

Ta lại có: x + y + z = 0

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

=> \(x^3+3x^2y+3xy^2+y^3=-z^3\)

=> \(x^3+y^3+z^3=-3x^2y-3xy^2\)

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

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

=> \(x^3+y^3+z^3=3xyz\)

Thay vào (*) ta có:

\(M=\frac{1}{2}\frac{3xyz}{xyz}=\frac{3}{2}\)

thay z = -(x+y) , y = -(z+x),... vao

=> Duoc bieu thuc trong do co 1/xy + 1/yz + 1/zx = (x+y+z)/xyz = 0