Ta có: x+y+z=0

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

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=0\)(1)

Ta có: \(K=\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)

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

\(=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2-x^2-y^2-z^2-2xy-2yz-2xz}\)

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

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

Vậy: \(K=\dfrac{1}{3}\)

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

\(K=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2}=\dfrac{1}{3}\)

x - y + z 2  +  z - y 2  + 2(x – y + z)(y – z)

=  x - y + z 2  + 2(x – y + z)(y – z) +  y - z 2

= x - y + z + y - z 2 = x 2

(x + y + z)2 – 2.(x + y + z).(x + y) + (x + y)2

= [(x + y + z) – (x + y)]2 (Áp dụng HĐT (2) với A = x + y + z ; B = x + y)

= z2.

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

Sai đề.

=(x-y+z)²  + 2.(x-y+z).(y-z)+ (y-z)²=(x-y+z+y-z)²=x²

(x-y+z)² + (z-y)² + 2.(x-y+z).(y-z)

= (x-y+z)² + (y-z)² + 2.(x-y+z).(y-z)

=[(x-y+z)+(y-z)]²

=(x-y+z+y-z)²

=x²

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

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

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

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

\(=x^2\)

hằng đẳng thức nha đổi vị trí tth]s 2 xuoong3 và 3 lên 2 ra rồi tự làm nha

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

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

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

= \(\left(x-y+z-z+y\right)^2\)

= \(x^2\)

(x-y+z)² + (z-y)² + 2(x-y+z).(y-z)

= (x-y+z)² + 2(x-y+z)(y-z) + (y-z)²

= (x-y+z+y-z)²

= x²