|x|=y-2003
|y|= z-2003
|z|=x-2003
Tìm x,y,z
tìm x,y,z :(1) |x|=y-2003 ; (2)|y|=z-2003;(3)|z|=x-2003
cho 1 tick, mình giải chi tiết cho, mình học dạng này rồi, dẽ cực lun, có gì lien hệ nah
tìm x,y,z :(1) |x|=y-2003 ; (2)|y|=z-2003;(3)|z|=x-2003
tìm x,y,z :(1) |x|=y-2003 ; (2)|y|=z-2003;(3)|z|=x-2003
|x|=y-2003
|y|= z-2003
|z|=x-2003
Tìm x,y,z
tìm x,y,z thuộc 3 điều kiện :(1) |x|=y-2003 ; (2)|y|=z-2003;(3)|z|=x-2003
C/m nếu \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
thì \(\dfrac{1}{x^{2003}}+\dfrac{1}{y^{2003}}+\dfrac{1}{z^{2003}}=\dfrac{1}{x^{2003}+y^{2003}+z^{2003}}\)
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)
Chứng minh :
Nếu \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\) thì \(\frac{1}{x^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{x^{2003}+y^{2003}+z^{2003}}\)
\(x;y;z\ne0\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\xy=-z\left(x+y+z\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-y\\xy+xz+yz+z^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\\left(x+z\right)\left(y+z\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\x=-z\end{matrix}\right.\)
- 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
CM rằng nếu: 1/x+1/y+1/z=1/x+y+z thì
1/x2003+1/y2003+1/z2003=1/x2003+y2003+z2003
Cố giúp mk câu này nha Hùng Hoàng!
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+xz}{xyz}=\frac{1}{x+y+z}\)
(x+y+z)(xy+yz+xz)=xyz
google seach
ta suy ra
(x+y)(y+z)(z+x)=0
\(x=-y\)
\(\frac{1}{x^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{-y^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{z^{2003}}\)
\(\frac{1}{x^{2003}+y^{2003}+z^{2003}}=\frac{1}{-y^{2003}+y^{2003}+z^{2003}}=\frac{1}{z^{2003}}\)
suy ra \(\frac{1}{x^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{x^{2003}+y^{2003}+z^{2003}}\)
Làm tương tự với các TH x= -z và y= -z
Từ đó ta được điều phải cm
C/m nếu \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\) thì \(\dfrac{1}{x^{2003}}+\dfrac{1}{y^{2003}}+\dfrac{1}{z^{2003}}=\dfrac{1}{x^{2003}+y^{2003}+^{2003}}\)
\(\dfrac{xy+xz+yz}{xyz}=\dfrac{1}{x+y+z}\)
\(\left(xy+xz+yz\right)\left(x+y+z\right)=xyz\)
\(x^2y+xy^2+xyz+x^2z+xyz+xz^2+xyz+y^2z+z^2y=xyz\)
\(x^2\left(y+z\right)+xy\left(y+z\right)+xz\left(z+y\right)+yz\left(y+z\right)=0\)
\(\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]=0\)
\(\left(y+z\right)\left(x+z\right)\left(x+y\right)=0\)
\(\left[{}\begin{matrix}x=-y\\z=-x\\y=-z\end{matrix}\right.\)
\(\dfrac{1}{x^{2003}}+\dfrac{1}{y^{2003}}+\dfrac{1}{z^{2003}}=\dfrac{1}{z^{2003}}=\dfrac{1}{x^{2003}+y^{2003}+z^{2003}}\)