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trả lời:
ta có:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\)
\(Q=\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\)
\(=\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}\)
\(=\left(\frac{x}{z}+\frac{x}{y}\right)+\left(\frac{y}{z}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{z}{y}\right)\)
\(=x\left(\frac{1}{z}+\frac{1}{y}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=x\left(-\frac{1}{x}\right)+y\left(-\frac{1}{y}\right)+z\left(-\frac{1}{z}\right)\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(=-3\)
~hok tốt~
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)=> (x+y+z)\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)=0
=> \(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}+3=0\)
=> \(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}=-3\)
Áp dụng BĐT Cauchy dạng Engel , ta được
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^3}{x+y+z}=\frac{1}{x+y+z}\)
Dấu "=" xảy ra khi \(\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\) => \(x=y=z\).(*)
Áp dụng BĐT Cauchy dạng Engel , ta được : \(\frac{1}{x^5}+\frac{1}{y^5}+\frac{1}{z^5}\ge\frac{\left(1+1+1\right)^3}{x^5+y^5+z^5}\) \(=\frac{1}{x^5+y^5+z^5}\)
Dấu "=" xảy ra khi x=y=z ( đã có ở (*) )
Vậy \(\frac{1}{x^5}+\frac{1}{y^5}+\frac{1}{z^5}=\frac{1}{x^5+y^5+z^5}\) ( đpcm) với x=y=z
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz--xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
CMTT:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
A=\(\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
\(A=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}\left(1\right)\)
mà \(xy+yz+xz=0\)
Từ \(\Rightarrow\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=0\)
Vậy A=0
\(\frac{1}{x}+\frac{1}{y}=5\)(1)
\(\frac{1}{x}-\frac{1}{y}=1\)(2)
(1)+(2)=> 1/x=3
(1)-(2)=>1/y=2
1/3+1/2=5/6
DS: 5/6