Tìm các số hữu tỉ x,y,z biết các số đó thõa mãn các điều kiện: xy= 1/3 yz= -2/5 xz= -3/10 10/07/2021 Bởi Jasmine Tìm các số hữu tỉ x,y,z biết các số đó thõa mãn các điều kiện: xy= 1/3 yz= -2/5 xz= -3/10
Ta có: `xy . yz . xz = 1/3 . (-2/5) . (-3/10)` `=> xyyzxz = 1/25` `=>x^2y^2z^2 =1/25` `=> (xyz)^2 = 1/25` `=>` \(\left[ \begin{array}{l}xyz = \frac{1}{5}\\xyz= -\frac{1}{5}\end{array} \right.\) +) Nếu `xyz= 1/5` Mà `xy = 1/3` `=> 1/3 z= 1/5` `=> z= 1/5 : 1/3` `=> z= 1/5 .3` `=> z= 3/5` `=> y= -2/5 : 3/5 = -2/5 . 5/3 = -2/3` `=> x = 1/3 : (-2/3) = 1/3 . (-3/2) = -1/2` +) Nếu `xyz= -1/5` Mà `xy= 1/3` `=> 1/3 z = -1/5` `=> z= -1/5 : 1/3` `=> z= -1/5. 3` `=> z= -3/5` `=> y= -2/5 : (-3/5)` `=> y= -2/5 . (-5/3)` `=>y= 2/3` `=> x= 1/3 : 2/3` `=> x= 1/3 . 3/2 = 1/2` Vậy `x = -1/2; y= -2/3 ; z= 3/5` hoặc `x= 1/2 ; y =2/3 ; z= -3/5` Bình luận
Theo đề ra , ta có : `x.y.y.z.x.z=1/3 . (-2/5).(-3/10)``=>.y.y.z.x.z=1/25``=>(x.y.z)²=1/25``=>` \(\left[ \begin{array}{l}x.y.z=\dfrac{1}{5}\\x.y.z=-\dfrac{1}{5}\end{array} \right.\) + Nếu `x.y.z=1/5` thì :`=>` \(\left[ \begin{array}{l}x=\dfrac{1}{5}:\dfrac{1}{3}\\y=\dfrac{1}{5}:(-\dfrac{2}{5})\\z=\dfrac{1}{5}:(-\dfrac{3}{10})\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=\dfrac{3}{5}\\y=-\dfrac{1}{2}\\z=-\dfrac{2}{3}\end{array} \right.\) + Nếu `x.y.z=-1/5` thì :`=>` \(\left[ \begin{array}{l}x=(-\dfrac{1}{5}):\dfrac{1}{3}\\y=(-\dfrac{1}{5}):(-\dfrac{2}{5})\\z=(-\dfrac{1}{5}):(-\dfrac{3}{10})\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=-\dfrac{3}{5}\\y=\dfrac{1}{2}\\z=\dfrac{2}{3}\end{array} \right.\) Vậy `(x;y;z)=(3/5:-1/2;-2/3),(-3/5;1/2;2/3)` Bình luận
Ta có: `xy . yz . xz = 1/3 . (-2/5) . (-3/10)`
`=> xyyzxz = 1/25`
`=>x^2y^2z^2 =1/25`
`=> (xyz)^2 = 1/25`
`=>` \(\left[ \begin{array}{l}xyz = \frac{1}{5}\\xyz= -\frac{1}{5}\end{array} \right.\)
+) Nếu `xyz= 1/5`
Mà `xy = 1/3`
`=> 1/3 z= 1/5`
`=> z= 1/5 : 1/3`
`=> z= 1/5 .3`
`=> z= 3/5`
`=> y= -2/5 : 3/5 = -2/5 . 5/3 = -2/3`
`=> x = 1/3 : (-2/3) = 1/3 . (-3/2) = -1/2`
+) Nếu `xyz= -1/5`
Mà `xy= 1/3`
`=> 1/3 z = -1/5`
`=> z= -1/5 : 1/3`
`=> z= -1/5. 3`
`=> z= -3/5`
`=> y= -2/5 : (-3/5)`
`=> y= -2/5 . (-5/3)`
`=>y= 2/3`
`=> x= 1/3 : 2/3`
`=> x= 1/3 . 3/2 = 1/2`
Vậy `x = -1/2; y= -2/3 ; z= 3/5` hoặc `x= 1/2 ; y =2/3 ; z= -3/5`
Theo đề ra , ta có :
`x.y.y.z.x.z=1/3 . (-2/5).(-3/10)`
`=>.y.y.z.x.z=1/25`
`=>(x.y.z)²=1/25`
`=>` \(\left[ \begin{array}{l}x.y.z=\dfrac{1}{5}\\x.y.z=-\dfrac{1}{5}\end{array} \right.\)
+ Nếu `x.y.z=1/5` thì :
`=>` \(\left[ \begin{array}{l}x=\dfrac{1}{5}:\dfrac{1}{3}\\y=\dfrac{1}{5}:(-\dfrac{2}{5})\\z=\dfrac{1}{5}:(-\dfrac{3}{10})\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=\dfrac{3}{5}\\y=-\dfrac{1}{2}\\z=-\dfrac{2}{3}\end{array} \right.\)
+ Nếu `x.y.z=-1/5` thì :
`=>` \(\left[ \begin{array}{l}x=(-\dfrac{1}{5}):\dfrac{1}{3}\\y=(-\dfrac{1}{5}):(-\dfrac{2}{5})\\z=(-\dfrac{1}{5}):(-\dfrac{3}{10})\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=-\dfrac{3}{5}\\y=\dfrac{1}{2}\\z=\dfrac{2}{3}\end{array} \right.\)
Vậy `(x;y;z)=(3/5:-1/2;-2/3),(-3/5;1/2;2/3)`