(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.

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

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

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

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

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

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

\(=-z^2\)

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: \(K=\dfrac{1}{3}\)

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

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

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.