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

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

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

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.

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: \(K=\dfrac{1}{3}\)

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

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

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

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

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

= (x + y + z + x +y)²

= (2x + 2y + z)²

Chúc bạn học tốt !

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

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

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

\(=z^2\)

Áp dụng BĐT: \(\left(a-b\right)^2=a^2-2ab+b^2\)