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\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz--xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
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
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)nhân lần lượt với x; y; z, ta có:
\(1+\frac{x}{y}+\frac{x}{z}=0\)(1)
\(1+\frac{y}{z}+\frac{y}{x}=0\)(2)
\(1+\frac{z}{x}+\frac{z}{y}=0\)(3)
Từ: (1); (2) và (3) => \(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=-3\)(*)
Mặt khác: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)quy đồng ta có:
\(\frac{\left(xy+yz+zx\right)}{xyz}=0\)hay xy + yz + zx = 0
Hay: \(\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right).\left(xy+yz+zx\right)=0\)
Khai triển, ta có:
\(\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{z}{x}+\frac{y}{x}+\frac{z}{y}=0\)
Vậy: \(\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=-\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)=3\)
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.
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow xy+yz+xz=0\) (nhân 2 vế với\(xyz\ne0\))
=> x2 + 2yz = x2 + 2yz - xy - yz - xz = x2 - xz - xy + yz = x(x - z) - y(x - z) = (x - y)(x - z).
Tương tự,y2 + 2xz = (y - x)(y - z) ; z2 + 2xy = (z - x)(z - y)
\(\Rightarrow\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}=1\)
c này mk tìm ra cách giải r nhé
_giải sao vậy bạn?