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Ta có : \(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}=\frac{x+y+z}{y+z+x}=1\) (do \(x+y+z\ne0\))

\(\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}}\)

\(\Leftrightarrow x=y=z\)

Thay \(x=y=z\) vào \(N=\frac{x^{123}.y^{456}}{z^{579}}\), ta có :

\(N=\frac{x^{123}.x^{456}}{x^{579}}\)

\(\Leftrightarrow\frac{x^{579}}{x^{579}}=1\)

Vậy N = 1

fai fai ối dồi ôi luôn

\(\Rightarrow\frac{x+1+y+2+z+3}{3+4+5}\)

\(\Rightarrow\frac{24}{12}=2\)

\(\frac{x+1}{3}=2\Rightarrow x=5\)

\(\frac{y+2}{4}=2\Rightarrow y=6\)

\(\frac{z+3}{5}=2\Rightarrow z=7\)

Áp dụng tính chất dãy tie số bằng nhau ta có:

\(\frac{x-y-z}{x}=\frac{y-z-x}{y}=\frac{z-x-y}{z}=\frac{x-y-z+y-z-x+z-x-y}{x+y+z}=-\frac{\left(x+y+z\right)}{x+y+z}=-1\)

\(\Rightarrow\hept{\begin{cases}x-y-z=-x\\y-z-x=-y\\z-y-x=-z\end{cases}\Rightarrow\hept{\begin{cases}y+z=-2x\\z+x=-2y\\x+y=-2z\end{cases}}}\)

\(\Rightarrow\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)=\frac{\left(x+y\right)}{x}.\frac{\left(y+z\right)}{y}.\frac{\left(z+x\right)}{z}=-\frac{8xyz}{xyz}=-8\)

1/x+1/y-1/z=(yz+xz-xy)/(xyz)=0 vì x,y,z#0 =>yz+xz-xy=0

x^2 + y^2 +z^2=(x+y-z)^2 +2(xz+yz-xy)=4

\(x+y+z=0\)

⇔\(-x=y+z\)

⇔\(x^2=\left(y+z\right)^2\) 

⇔\(x^2=y^2+2yz+z^2\) 

⇔\(y^2+z^2-x^2=-2yz\)

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

x+y+z=-3 => (x+1)+(y+1)+(z+1)=0

Đặt x+1=a,y+1=b,z+1=c ta có:

a+b+c=0 => a³+b³+c³=3abc (tự cm) hay (x+1)³+(y+1)³+(z+1)³=3(x+1)(y+1)(z+1) (dpdcm)