Toán cho 1/x+1/y+1/z= 1/(x+y+z) cmr: 1/x^2019+1/y^2019+1/z^2019=1/(x^2019+y^2019+z^2019) 01/12/2021 By Piper cho 1/x+1/y+1/z= 1/(x+y+z) cmr: 1/x^2019+1/y^2019+1/z^2019=1/(x^2019+y^2019+z^2019)
Đáp án: Ta có: `1/x+1/y+1/z=1/(x+y+z)` `=> 1/x+1/y=1/(x+y+z)-1/z` `=> (x+y)/(xy)=-(x+y)/(z.(x+y+z))` `=> (x+y).[z(x+y+z)+xy]=0 ` `=> (x+y)(y+z)(z+x)=0` `=>` \(\left[ \begin{array}{l}x+y=0\\y+z=0\\z+x=0\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=-y\\y=-z\\z=-x\end{array} \right.\) `=> 1/(x^2019)+1/(y^2019)+1/(z^2019)=1/(z^2019)` `1/(x^2019+y^2019+z^2019)=1/(z^2019)` `=> 1/(x^2019)+1/(y^2019)+1/(z^2019)=1/(x^2019+y^2019+z^2019)=1/(z^2019)` `=> đpcm` Trả lời
Có : $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} = \dfrac{1}{x+y+z}$ $\to \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x+y+z}-\dfrac{1}{z}$ $\to \dfrac{x+y}{xy} = \dfrac{-(x+y)}{z.(x+y+z)}$ $\to (x+y).[z.(x+y+z)+xy] = 0 $ $\to (x+y).(y+z).(z+x) = 0 $ $\to$ Trong ba số $x,y,z$ luôn có hai số đối nhau. Không mất tính tổng quát giả sử $x=-y$ Khi đó : $\dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} = \dfrac{1}{z^{2019}}$ Và $\dfrac{1}{x^{2019}+y^{2019}+z^{2019}} = \dfrac{1}{z^{2019}}$ Do đó : $\dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} = \dfrac{1}{x^{2019}+y^{2019}+z^{2019}} $ Trả lời
Đáp án:
Ta có: `1/x+1/y+1/z=1/(x+y+z)`
`=> 1/x+1/y=1/(x+y+z)-1/z`
`=> (x+y)/(xy)=-(x+y)/(z.(x+y+z))`
`=> (x+y).[z(x+y+z)+xy]=0 `
`=> (x+y)(y+z)(z+x)=0`
`=>` \(\left[ \begin{array}{l}x+y=0\\y+z=0\\z+x=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=-y\\y=-z\\z=-x\end{array} \right.\)
`=> 1/(x^2019)+1/(y^2019)+1/(z^2019)=1/(z^2019)`
`1/(x^2019+y^2019+z^2019)=1/(z^2019)`
`=> 1/(x^2019)+1/(y^2019)+1/(z^2019)=1/(x^2019+y^2019+z^2019)=1/(z^2019)`
`=> đpcm`
Có : $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} = \dfrac{1}{x+y+z}$
$\to \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x+y+z}-\dfrac{1}{z}$
$\to \dfrac{x+y}{xy} = \dfrac{-(x+y)}{z.(x+y+z)}$
$\to (x+y).[z.(x+y+z)+xy] = 0 $
$\to (x+y).(y+z).(z+x) = 0 $
$\to$ Trong ba số $x,y,z$ luôn có hai số đối nhau.
Không mất tính tổng quát giả sử $x=-y$
Khi đó : $\dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} = \dfrac{1}{z^{2019}}$
Và $\dfrac{1}{x^{2019}+y^{2019}+z^{2019}} = \dfrac{1}{z^{2019}}$
Do đó : $\dfrac{1}{x^{2019}} + \dfrac{1}{y^{2019}} + \dfrac{1}{z^{2019}} = \dfrac{1}{x^{2019}+y^{2019}+z^{2019}} $