Ta có: \(\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3\)

\(=\left(x^2-y^2+y^2-z^2\right)^3-3\left(x^2-y^2+y^2-z^2\right)\left(x^2-y^2\right)\left(y^2-z^2\right)+\left(z^2-x^2\right)^3\)

\(=\left(x^2-z^2\right)^3-3\left(x^2-z^2\right)\left(x^2-y^2\right)\left(y^2-z^2\right)+\left(z^2-x^2\right)^3\)

\(=-3\left(x^2-y^2\right)\left(x^2-z^2\right)\left(y^2-z^2\right)\)

=-3(x-y)(x+y)(x-z)(x+z)(y-z)(y+z)

Ta có: \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

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

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

=-3(x-y)(y-z)(x-z)

Ta có: \(C=\frac{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}{\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3}\)

\(=\frac{-3(x-y)(x+y)(x-z)(x+z)(y-z)(y+z)}{-3\left(x-y\right)\left(y-z\right)\left(x-z\right)}=\left(x+y\right)\left(x+z\right)\left(y+z\right)\)

Ta có (a - b)² = (b - a)² 

Vậy biểu thức viết lại dưới dạng: a² + 2ab + b² (Với a = x - y + z và b = y - z) 
(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

a: Sửa đề: \(\left(x-z\right)^2+z\left(x-z\right)\left(x+z\right)+\left(x+z\right)^2\)

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

\(=2x^2+2z^2+x^2z-z^3\)

b: \(\left(x-y\right)^2+\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)

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

\(=3x^2+y^2\)

giup mik nha tí 30p nữa mình on cam on mn

Hình như ghi sai đề hay sao í

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.