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Với x,y,z khác 0 ta có \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=>\frac{yz+xz+xy}{xyz}=0=>yz+xz+xy=0\)
Ta luôn có nếu a+b+c=0 thì a3+b3+c3=3abc
Vì xy+yz+zx=0 nên x3y3+y3z3+z3x3=3x2y2z2
Với x3y3+y3z3+z3x3=3x2y2z2 ta có:
\(\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{y^3z^3+x^3z^3+x^3y^3}{x^2y^2z^2}=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)
Vậy ....
Ta có 1/x+1/y+1/z=0
=>1/x+1/y=-1/z
=>(1/x+1/y)^3= (-1/z)^3
=>1/x^3+1/y^3+3.1/x.1/y.(1/x+1/y) =-1/z^3
=>1/x^3+1/y^3+1/z^3= -3.1/x.1/y.(1/x+1/y) =3/(xyz) (vì 1/x+1/y=-1/z)
Mặt khác: 1/x+1/y+1/z=0
=>(xy+yz+zx)/(xyz)=0
=>xy+yz+zx=0
A=yz/x^2 +2yz + xz/y^2+ 2xz + xy/z^2+ 2 xy
=xyz/x^3+xyz/y^3+xyz/z^3 +2(xy+yz+zx) (vì x,y,z khác 0)
=xyz(1/x^3+1/y^3+1/z^3) (vì xy+yz+zx=0)
=xyz.3/(xyz) (vì 1/x^3+1/y^3+1/z^3=3/(xyz) )
=3
Vậy A=3.
Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\hept{\begin{cases}1+\frac{x}{y}+\frac{x}{z}=0\\\frac{y}{x}+1+\frac{y}{z}=0\\\frac{z}{x}+\frac{z}{y}+1=0\end{cases}}\)
\(\Rightarrow\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}=-3\)
mà \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{yz+xz+xy}{xyz}=0\)
\(\Rightarrow yz+xz+xy=0\)
\(\Rightarrow\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\left(yz+xz+xy\right)=0\)
\(\Rightarrow\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}=0\)
\(\Rightarrow\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=3\)
\(\Rightarrow\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{xz}{y^2}=3\)
Học tốt
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
<=> \(\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
<=> \(\left(\frac{1}{x}+\frac{1}{y}\right)^3=\left(-\frac{1}{z}\right)^3\)
<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{x^2y}+\frac{3}{xy^2}=-\frac{1}{z^3}\)
<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)\)
<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-\frac{3}{xy}.\left(-\frac{1}{z}\right)\)
<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
Khi đó: P = \(\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{xz}{y^2}=\frac{xyz}{z^3}+\frac{xyz}{x^3}+\frac{xyz}{y^3}=xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz\cdot\frac{3}{xyz}=3\)