\(x+y+z=0\)

\(\Rightarrow x+y=-z\)

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

\(\Rightarrow x^2+2xy+y^2=z^2\)

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

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

\(\Rightarrow x^4+y^4+z^4+2x^2y^2-2y^2z^2-2x^2z^2=4x^2y^2\)

\(\Rightarrow x^4+y^4+z^4=4x^2y^2-2x^2y^2+2y^2z^2+2x^2z^2\)

\(\Rightarrow x^4+y^4+z^4=2x^2y^2+2y^2z^2+2x^2z^2\)

\(\Rightarrow2\left(x^4+y^4+z^4\right)=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2x^2z^2\)

\(\Rightarrow2\left(x^4+y^4+z^4\right)=\left(x^2+y^2+z^2\right)^2\)

\(\Rightarrow x^4+y^4+z^4=\frac{\left(x^2+y^2+z^2\right)^2}{2}\)

a, Với \(x^2+y^2+z^2=2\)thì \(x^4+y^4+z^4=\frac{2^2}{2}=2\)

b, Với \(x^2+y^2+z^2=1\)thì \(x^4+y^4+z^4=\frac{1^2}{2}=\frac{1}{2}\)

Chúc bạn học tốt.

ta có: a² + b² = 2ab

=> a² + b² - 2ab = 0

=> (a-b)² = 0

=> a - b = 0

=> a = b

a) \(-3x^2+27=0\)

\(\Leftrightarrow-3\left(x^2-9\right)=0\)

\(\Leftrightarrow-3\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}}\)

b)\(2x^3+54=0\)

\(\Leftrightarrow2\left(x^3+27\right)=0\)

\(\Leftrightarrow2\left(x+3\right)\left(x^2-3x+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2-x+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x\left(x-1\right)=-9\left(loại\right)\end{cases}}}\)

ý a bạn giải kiểu gì?

 Ta có:

\(I=y^2+4x+5x^2-2xy\)

\(I=4x^2+4x+1+x^2-2xy+y^2-1\)

\(I=\left(2x+1\right)^2+\left(x-y\right)^2-1\)

Mà: \(\hept{\begin{cases}\left(2x+1\right)^2\ge0\\\left(x-y\right)^2\ge0\end{cases}\Rightarrow\left(2x+1\right)^2+\left(x-y\right)^2\ge0\left(\forall x,y\in R\right)}\)

\(\Rightarrow I=\left(2x+1\right)^2+\left(x-y\right)^2-1\ge-1\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}2x+1=0\\x-y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=y\end{cases}\Leftrightarrow}x=y=-\frac{1}{2}}\)

Vậy Min I = -1 khi x = y = -1/2

Bài này không suy ra được GTLN nha bạn