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

<=>   \(2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

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

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

<=>  \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\) <=>  \(\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}}\)<=>  \(x=y=z\)  (đpcm)

      \(x^2+y^2+z^2=xy+xz+yz\)

\(\Rightarrow2x^2+2y^2+2z^2=2xy+2xz+2yz\)

\(\Rightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

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

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)(1)

Ta có: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x;y\\\left(y-z\right)^2\ge0\forall y;z\\\left(x-z\right)^2\ge0\forall x;z\end{cases}\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x;y;z}\) (2)

Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}\Rightarrow}x=y=z}\)

Chúc bạn học tốt.

\(\left(x-2\right)^2-\left(x+3\right)^2+\left(x+4\right)\left(x-4\right)\)

\(=x^2-4x+4-x^2-6x-9+x^2-16\)

\(=x^2-10x-21\)

do k biết đề nên mk rút gọn

Sửa đề chút: a+b+c=0

Ta có: \(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^2=0\)

\(a^2+b^2+c^2+2ab+2bc+2ca=0\)

\(\Rightarrow2\left(ab+bc+ca\right)=0-1\)

\(\Rightarrow ab+bc+ca=-\frac{1}{2}\)

\(\Rightarrow\left(ab+bc+ca\right)^2=\frac{1}{4}\)

\(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab^2c+2bc^2a+2ca^2b=\frac{1}{4}\)

\(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc.\left(a+b+c\right)=\frac{1}{4}\)

\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\frac{1}{4}\)

Ta có: \(a^2+b^2+c^2=1\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)

\(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=1\)

\(a^4+b^4+c^4+2.\left[\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\right]=1\)

\(a^4+b^4+c^4+2.\frac{1}{4}=1\)

\(a^4+b^4+c^4+\frac{1}{2}=1\)

\(\Rightarrow a^4+b^4+c^4=\frac{1}{2}\)

Vậy \(a^4+b^4+c^4=\frac{1}{2}\)

Nếu mk sửa đề sai thì bảo mk nhé.( mk lm đúng để của b thử rồi nhưng ko ra)

bn sửa đề thì mk cx ra rồi nhưng quan trọng nó là 1 chứ ko phải là 0