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\(x+y+z=2018\)\(\Rightarrow\)\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2018}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\\ \Leftrightarrow x^2y+xy^2+xyz+xyz+y^2z+\\ yz^2+zx^2+xyz+z^2x-xyz=0\)
\(\Leftrightarrow x^2y+xy^2+xyz+xyz+\\ y^2z+yz^2+zx^2+z^2x=0\)
\(\Leftrightarrow xy\left(x+y\right)+yz\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)=0\\ \Leftrightarrow\left(x+y\right)\left(xy+yz+xz+z^2\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y\left(x+z\right)+z\left(x+z\right)\right)=0\\ \Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
suy ra x+y=0 hoặc y+z=0 hoặc x+z=0
hay x=-y hoặc y=-z hoặc x=-z
thay vào D ta tính dc kq
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 \(\dfrac{1}{x^{2007}}+\dfrac{1}{y^{2007}}+\dfrac{1}{z^{2007}}=\dfrac{1}{x^{2007}+y^{2007}+z^{2007}}\).
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
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{xy}{z}+\dfrac{yz}{x}\) ≥ \(2\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}=2\sqrt{y^2}=2y\left(1\right)\)
\(\dfrac{yz}{x}+\dfrac{xz}{y}\) ≥ \(2\sqrt{\dfrac{yz}{x}.\dfrac{xz}{y}}=2\sqrt{z^2}=2z\left(2\right)\)
\(\dfrac{xy}{z}+\dfrac{xz}{y}\) ≥ \(2\sqrt{\dfrac{xy}{z}.\dfrac{xz}{y}}=2\sqrt{x^2}=2x\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3) , ta được :
\(2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\right)\) ≥ \(2\left(x+y+z\right)\)
⇔ \(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\) ≥ \(x+y+z=2019\)
⇒ \(P_{Min}=2019\) ⇔ \(x=y=z=673\)
\(A=\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{z+x}{y}\) (đã sửa đề)
\(A+3=\dfrac{x+y+z}{z}+\dfrac{x+y+z}{x}+\dfrac{x+y+z}{y}\)
\(A+3=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=0\)
\(A=-3\)
Do \(x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}\)
=> \(\left\{{}\begin{matrix}x+\dfrac{1}{y}=y+\dfrac{1}{z}\Leftrightarrow x-y=\dfrac{1}{z}-\dfrac{1}{y}\Leftrightarrow x-y=\dfrac{y-z}{yz}\\y+\dfrac{1}{z}=z+\dfrac{1}{x}\Leftrightarrow y-z=\dfrac{1}{x}-\dfrac{1}{z}\Leftrightarrow y-z=\dfrac{z-x}{xz}\\z+\dfrac{1}{x}=x+\dfrac{1}{y}\Leftrightarrow z-x=\dfrac{1}{y}-\dfrac{1}{x}\Leftrightarrow z-x=\dfrac{x-y}{xy}\end{matrix}\right.\)
=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=\dfrac{\left(y-z\right)\left(z-x\right)\left(x-y\right)}{x^2y^2z^2}\)
<=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)x^2y^2z^2=\left(y-z\right)\left(z-x\right)\left(x-y\right)\)
<=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x^2y^2z^2-1\right)=0\)
=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=0\) hoặc \(x^2y^2z^2-1=0\)
=> x=y=z hoặc xyz=1 hoặc xyz=-1