Ta có:      \(x+y+z=0\)

\(\Leftrightarrow\)  \(\left(x+y+z\right)^2=0\)

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

\(\Leftrightarrow\)\(x^2+y^2+z^2=0\)   (vì  xy + yz + xz =0)

\(\Leftrightarrow\)\(x=y=z=0\)

Vậy      \(S=\left(0-1\right)^{1999}+0^{2003}+\left(0+1\right)^{2006}=0\)

Sửa đề \(x^4+y^4+z^4=\frac{1}{2}\left(x^2+y^2+z^2\right)^2\)

Ta có: \(x+y+z=0\)

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

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

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

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

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

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

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

\(\Leftrightarrow2\left(x^4+y^4+z^4\right)=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2z^2x^2=\left(x^2+y^2+z^2\right)^2\)

\(\Leftrightarrow x^4+y^4+z^4=\frac{1}{2}\left(x^2+y^2+z^2\right)^2\left(đpcm\right)\)

Do z > 0 nên từ xy ² z ² + x ² z + y = 3^{z 2} ⇒ xy ² +\(\frac{x^2}{z}+\frac{y}{z^2}=3\)

Áp dụng AM­GM ta có:

(x ²y ² + y ² ) + (x ² +\(\frac{x^2}{z^2}\))+(\(\frac{y^2}{z^2}+\frac{1}{z^2}\)) ≥ 2(xy ² +\(\frac{x^2}{z}+\frac{y}{z^2}\))=6

...............

ai giúp mik vs huhu

2.Câu hỏi của Lãnh Hàn Thần - Toán lớp 8 - Học toán với OnlineMath

 Ta có: (x + y + z)^2 = 0 <=> x^2 + y^2 + z^2 + 2.(xy + yz + xz) = 0 
<=> 1 + 2(xy + yz + xz) = 0 
<=> xy + yz + xz = -1/2 
Lại có: x^2.y^2 + y^2.z^2 + x^2.z^2 = (xy + yz + xz)^2 - 2.xyz.(x + y + z) = 1/4 - 0 = 1/4 
=> x^4+y^4+z^4 = (x^2 + y^2 + z^2)^2 - 2.(x^2.y^2 + y^2.z^2 + x^2.z^2) = 1 - 2.1/4 = 1/2 
Vậy x^4 + y^4 + z^4 = 1/2

not think