=0 nha

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

2) \(x=y+1\Rightarrow x-y=1\)

\(\Rightarrow\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)=x^8-y^8\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)=x^8-y^8\)

\(\Leftrightarrow\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)=x^8-y^8\)

\(\Leftrightarrow\left(x^4-y^4\right)\left(x^4+y^4\right)=x^8-y^8\)

\(\Leftrightarrow x^8-y^8=x^8-y^8\)(đúng)

Vậy \(\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)=x^8-y^8\)(đpcm)

\(x+y+z=0\)

⇔\(-x=y+z\)

⇔\(x^2=\left(y+z\right)^2\) 

⇔\(x^2=y^2+2yz+z^2\) 

⇔\(y^2+z^2-x^2=-2yz\)

Tương tự:

\(z^2+x^2-y^2=-2zx\)

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

➞ S = \(\dfrac{1}{-2xy}+\dfrac{1}{-2yz}+\dfrac{1}{-2zx}=\dfrac{x+y+z}{-2xyz}=0\) 

Vậy S = 0

Ta có:

\(x+y+z=0\)

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

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

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

Tương tự ta được:
\(S=\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}=-\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=-\frac{1}{2}\cdot\frac{x+y+z}{xyz}=0\)

Vậy S=0

1,

\(x^2+y^2+y^2=14\)

\(\Rightarrow\left(x+y+z\right)^2-2xy-2yz-2zx=14\)

\(\Rightarrow-2\left(xy+yz+zx\right)=14\)

\(\Rightarrow xy+yz+zx=-7\)

\(\Rightarrow\left(xy+yz+zx\right)^2=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2x^2yz+2xy^2z+2xyz^2=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2xyz\left(x+y+z\right)=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2=49\)

Ta có: \(x^4+y^4+z^4\)

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

\(=14^2-2\left(x^2y^2+y^2z^2+z^2x^2\right)\)

\(=14^2-2.49\)

\(=196-98\)

\(=98\)