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Chào bạn

bạn nhân chéo lên rồi tách ra thì bạn sẽ có

1/x+1/y+1/z=1/x+y+z tương đương với (x+y)(y+z)(x+z)=0

Đến đây thì dễ rồi

\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{1}\)

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

\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{1}=\dfrac{x+y+z}{\left(y+z+1\right)+\left(x+z+1\right)+\left(x+y-2\right)}\)

\(=\dfrac{x+y+z}{2x+2y+2z}\)

\(TH1:x+y+z=0\)

⇒ \(\dfrac{x+y+z}{1}=0\)

⇒ \(x=y=z=0\)(loại vì trái với điều kiện đề bài )

\(TH2:z+y+z\)≠ 0

⇒ \(\dfrac{x+y+z}{2x+2y+2z}=\dfrac{x+y+z}{2.\left(x+y+z\right)}=\dfrac{1}{2}\)

Vậy \(x+y+z=\dfrac{1}{2}\)

\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\) ⇒ \(2x=y+z+1\)⇒\(2x=y+z+2\left(x+y+z\right)=2x+3y+3z\)

⇒ \(3y+3z=0\) ⇒ \(y+z=0\) ⇒ \(2x=1\) ⇒ \(x=\dfrac{1}{2}\)

\(\dfrac{y}{x+z+1}=\dfrac{1}{2}\) ⇒ \(2y=x+z+1=x+z+2\left(x+y+z\right)=2y+3x+3z\)

⇒ \(3x+3z=0\) ⇒ \(x+z=0\) ⇒ \(2y=1\) ⇒ \(y=\dfrac{1}{2}\)

\(x+z=0\) ; \(x=\dfrac{1}{2}\)

⇒ \(z=0-\dfrac{1}{2}=\dfrac{-1}{2}\)

Vậy \(x=\dfrac{1}{2}\) ; \(y=\dfrac{1}{2}\) ; \(z=\dfrac{-1}{2}\)

Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

\(\Leftrightarrow\dfrac{xy+yz+xz}{xyz}=\dfrac{1}{x+y+z}\)

\(\Leftrightarrow\left(xy+yz+xz\right)\left(x+y+z\right)=xyz\)

\(\Leftrightarrow x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+3xyz-xyz=0\)

\(\Leftrightarrow x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+2xyz=0\)

\(\Leftrightarrow x^2y+xy^2+x^2z+xyz+y^2z+yz^2+xz^2+xyz=0\)

\(\Leftrightarrow x\left(xy+y^2+xz+yz\right)+z\left(y^2+yz+xz+xy\right)=0\)

\(\Leftrightarrow x\left[y\left(x+y\right)+z\left(x+y\right)\right]+z\left[y\left(y+z\right)+x\left(y+z\right)\right]=0\)

\(\Leftrightarrow x\left(x+y\right)\left(y+z\right)+z\left(y+z\right)\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

* x = -y

\(\dfrac{1}{x^{2007}}+\dfrac{1}{y^{2007}}+\dfrac{1}{z^{2007}}=\dfrac{1}{x^{2007}}-\dfrac{1}{x^{2007}}+\dfrac{1}{z^{2007}}=\dfrac{1}{z^{2007}}\)(*)

\(\dfrac{1}{x^{2007}+y^{2007}+z^{2007}}=\dfrac{1}{x^{2007}-x^{2007}+z^{2007}}=\dfrac{1}{z^{2007}}\)(*)

Từ (*) và (**) \(\Rightarrow\) đpcm

Tương tự xét y = -z và z = -x

Vậy nếu x, y, z khác 0 và x + y +z khác 0 thì \(\dfrac{1}{x^{2007}}+\dfrac{1}{y^{2007}}+\dfrac{1}{z^{2007}}=\dfrac{1}{x^{2007}+y^{2007}+z^{2007}}\).

Lời giải:
Nếu $x+y+z=0$ thì:

$\frac{x+y-z}{z}=\frac{-z-z}{z}=-2$

$\frac{y+z-x}{x}=\frac{-x-x}{x}=-2$

$\frac{z+x-y}{y}=\frac{-y-y}{y}=-2$ 

(thỏa mãn đkđb)

Khi đó:

$P=(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=\frac{(x+y)(y+z)(z+x)}{xyz}$

$=\frac{(-z)(-x)(-y)}{xyz}=\frac{-xyz}{xyz}=-1$

Nếu $x+y+z\neq 0$

Áp dụng TCDTSBN:

$\frac{x+y-z}{z}=\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z+y+z-x+z+x-y}{z+x+y}=\frac{x+y+z}{x+y+z}=1$

$\Rightarrow x+y=2z; y+z=2x, z+x=2y$. Khi đó:

$P=\frac{(x+y)(y+z)(z+x)}{xyz}=\frac{2z.2x.2y}{xyz}=8$

Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)

\(\Rightarrow xy=-yz-xz;yz=-xy-xz;xz=-xy-yz\)

Ta lại có: \(A=\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}=\dfrac{x^2+xy}{xz}+\dfrac{z^2+xz}{yz}+\dfrac{y^2+yz}{xy}\)

\(=\dfrac{x^2-yz-xz}{xz}+\dfrac{z^2-xy-yz}{yz}+\dfrac{y^2-xy-xz}{xy}\)

\(=\dfrac{x\left(x-z\right)}{xz}-\dfrac{yz}{xz}+\dfrac{z\left(z-y\right)}{yz}-\dfrac{xy}{yz}+\dfrac{y\left(y-x\right)}{xy}-\dfrac{xz}{xy}\)

\(=\dfrac{x-z}{z}-\dfrac{y}{x}+\dfrac{z-y}{y}-\dfrac{x}{z}+\dfrac{y-x}{x}-\dfrac{z}{y}\)

\(=\dfrac{x-z-x}{z}+\dfrac{z-y-z}{y}+\dfrac{y-x-y}{x}=\dfrac{-z}{z}+\dfrac{-y}{y}+\dfrac{-x}{x}\)

\(=-1-1-1=-3\). Vậy A=-3

từ đề bài ta có bất đẳng thức cần chứng minh tương đương: 

\(3+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{9}{4}\)

<=>\(\dfrac{3}{4}+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

ta có \(\dfrac{3}{4}+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{3}{4}+\dfrac{z+y}{4x}+\dfrac{x+z}{4y}+\dfrac{x+y}{4z}=\dfrac{3}{4}+\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{4}=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(đpcm\right)\)Dấu "=" xảy ra khi x=y=z=\(\dfrac{1}{3}\)