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\(\dfrac{xy+xz+yz}{xyz}=\dfrac{1}{x+y+z}\)
\(\left(xy+xz+yz\right)\left(x+y+z\right)=xyz\)
\(x^2y+xy^2+xyz+x^2z+xyz+xz^2+xyz+y^2z+z^2y=xyz\)
\(x^2\left(y+z\right)+xy\left(y+z\right)+xz\left(z+y\right)+yz\left(y+z\right)=0\)
\(\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]=0\)
\(\left(y+z\right)\left(x+z\right)\left(x+y\right)=0\)
\(\left[{}\begin{matrix}x=-y\\z=-x\\y=-z\end{matrix}\right.\)
\(\dfrac{1}{x^{2003}}+\dfrac{1}{y^{2003}}+\dfrac{1}{z^{2003}}=\dfrac{1}{z^{2003}}=\dfrac{1}{x^{2003}+y^{2003}+z^{2003}}\)
\(x;y;z\ne0\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\xy=-z\left(x+y+z\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-y\\xy+xz+yz+z^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\\left(x+z\right)\left(y+z\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\x=-z\end{matrix}\right.\)
- Với \(x=-y\Rightarrow\frac{1}{x^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{-y^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{z^{2003}}\)
\(\frac{1}{x^{2003}+y^{2003}+z^{2003}}=\frac{1}{-y^{2003}+y^{2003}+z^{2003}}=\frac{1}{z^{2003}}\)
\(\Rightarrow\frac{1}{x^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{x^{2003}+y^{2003}+z^{2003}}\)
2 trường hợp còn lại tương tự
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
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 \(\dfrac{1}{x^{2007}}+\dfrac{1}{y^{2007}}+\dfrac{1}{z^{2007}}=\dfrac{1}{x^{2007}+y^{2007}+z^{2007}}\).
Bài này trên diễn đàn có nhiều thực chưa có bài thực sự đúng
\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\) (1)
đk: \(\left\{{}\begin{matrix}x+y\ne0\\x+z\ne0\\y+z\ne0\end{matrix}\right.\) Nếu x+y+z=0\(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)(*)
Thay (*) vào (1)
\(\dfrac{x}{-x}+\dfrac{y}{-y}+\dfrac{z}{-z}=-3\) kết luận: \(x+y+z\ne0\)
Nhân 2 vế (1) với x+y+z khác 0 ta có\(\left(1\right)\Leftrightarrow\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)\left(x+y+z\right)=\left(x+y+z\right)\)
\(\Leftrightarrow\left(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\right)+\left(y+z\right).\dfrac{y}{x+z}+\left(x+y\right).\dfrac{z}{x+y}+\left(x+z\right)\dfrac{x}{y+z}=\left(x+y+z\right)\)
\(\Leftrightarrow\left(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\right)+\left(x+y+z\right)=\left(x+y+z\right)\)\(\Rightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0\)
Vẫn lỗi:
\(.....\\ \left(x+z\right)\dfrac{x}{y+z}+\left(z+x\right)\dfrac{y}{z+x}+\left(x+y\right)\dfrac{z}{x+y}\)
....
Lời giải:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0\)
\(\Leftrightarrow (x+y)\left(\frac{1}{xy}+\frac{1}{z(x+y+z)}\right)=0\)
\(\Leftrightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0\)
\(\Leftrightarrow (x+y).\frac{z(y+z)+x(z+y)}{xyz(x+y+z)}=0\)
\(\Leftrightarrow \frac{(x+y)(z+x)(z+y)}{xyz(x+y+z)}=0\Rightarrow (x+y)(y+z)(x+z)=0\)
\(\Rightarrow \left[\begin{matrix} x=-y\\ y=-z\\ z=-x\end{matrix}\right.\)
Không mất tổng quát, giả sử \(x=-y\):
\(\frac{1}{x^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{(-y)^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{z^{2003}}\)
\(\frac{1}{x^{2003}+y^{2003}+z^{2003}}=\frac{1}{(-y)^{2003}+y^{2003}+z^{2003}}=\frac{1}{z^{2003}}\)
Do đó: \(\frac{1}{x^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{x^{2003}+y^{2003}+z^{2003}}\) (đpcm)