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

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

Mik mới biết làm câu a thôi còn câu b thì từ từ mik nghĩ đã nhé @-@

Chúc bn học giỏi nhoa!!!

Thanh Hà cảm ơn bạn nha 

Ta có: \(x+y+z=0\)

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

\(x^2+y^2+z^2+2xy+2yz+2xz=0\)

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

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

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

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

\(=\frac{-2.\left(xy+yz+zx\right)}{2.\left[-2.\left(xy+yz+zx\right)\right]-2.\left(xy+yz+zx\right)}\)

\(=\frac{-2.\left(xy+yz+zx\right)}{-6.\left(xy+yz+zx\right)}\)

\(=\frac{1}{3}\left(xy+yz+zx\ne0\right)\)

Tham khảo nhé~

\(x+y+z=0\)

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

\(x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)

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

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

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

\(=\frac{-2\left(xy+yz+zx\right)}{2\left[-2\left(xy+yz+zx\right)\right]-2\left(xy+yz+xz\right)}\)

\(=\frac{-2\left(xy+yz+zx\right)}{-4\left(xy+yz+zx\right)-2\left(xy+yz+xz\right)}\)

\(=\frac{-2\left(xy+yz+zx\right)}{-6\left(xy+yz+zx\right)}\)

\(=\frac{1}{3}\)

Ta có: \(x+y+z=0\)

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

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

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

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

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

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

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

\(=z\)

Ta có: \(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\x+z=-y\\y+z=-x\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2=z^2\\\left(x+z\right)^2=y^2\\\left(y+z\right)^2=x^2\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+2xy+y^2=z^2\\x^2+2xz+z^2=y^2\\y^2+2yz+z^2=x^2\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2+y^2-z^2=-2xy\\x^2+z^2-y^2=-2xz\\y^2+z^2-x^2=-2yz\end{cases}}\)

\(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xy^2}\)

\(=\frac{\left(-2xy\right).\left(-2yz\right).\left(-2xz\right)}{16xy^2}\)

\(=\frac{-8x^2y^2z^2}{16xy^2}\)

\(=\frac{-xz^2}{2}\left(x,y\ne0\right)\)

Ta có \(x+y+z=0\Rightarrow x+y=-z\Rightarrow x-y=z\Rightarrow\left(x-y\right)^2=z^2\)

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

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

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

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

\(=1\)

Vậy \(x+y+z=0\)thì \(A=1\)

Cho mình hỏi tại sao x + y = -z \(\Rightarrow\)x-y = z

x^2+y^2+z^2/y^2-2yx+z^2+z^2-2xy+x^2+x^2-2xy+y^2=x^2+y^2+z^2/2y^2+2x^2+2z^2-6xy=x^2+y^2+z^2/2(x^2+y^2+z^2)-6xy=1/2-6xy

xét mẫu ta có

=y^2 - 2yz + z^2 + z^2 -2xz + x^2 + x^2 -2xy +y^2

thêm bớt  x^2,y^2,z^2 vào mẫu ta có

=3y^2 + 3x^2 + 3z^2 - (x^2 + y^2 + z^2 + 2xy + 2yz + 2xz)

đúng không

mà (x+y+z)=0 => (x+y+z)^2=0

mà (x^2 + y^2 + z^2 + 2xy + 2yz + 2xz) phân tích ra thành (x+y+z)^2

=> (x^2 + y^2 + z^2 + 2xy + 2yz + 2xz)=0

=> (x^2 + y^2 + z^2 )/ 3(x^2 + y^2 + z^2)

rút gọn thành 1/3

nhớ k nha chuẩn 100%

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

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

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

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

\(=\frac{-2\left(xy+yz+zx\right)}{2\left[-3\left(xy+yz+zx\right)\right]}=\frac{1}{3}\)