Tìm x,y,z; biết: ( x- 1/5)( y + 1/2)( z-3)=0 x+1=y+2=z+3 Giúp mik với mik dg cần gấp

`( x- 1/5)( y + 1/2)( z-3)=0`

`⇒` \(\left[ \begin{array}{l}x=\dfrac{1}{5}\\y=-\dfrac{1}{2}\\z=3\end{array} \right.\) 

$TH1$. $x = \dfrac{1}{5}$; $y;z ∈ R$

$⇒$ $\dfrac{1}{5} + 1 = y + 2 = z+3$

$⇒$ $\left\{\begin{matrix}y = -\dfrac{4}{5} & \\ z = -\dfrac{9}{5}& \end{matrix}\right.$

$TH2$. $y= -\dfrac{1}{2} ; x;z ∈ R$

$⇒ x + 1 = \dfrac{-1}{2} + 2 = z+3$

$⇒$ $\left\{\begin{matrix}x = \dfrac{1}{2} & \\ z = -\dfrac{3}{2}& \end{matrix}\right.$

$TH3$. $z = 3; x;y ∈ R$

$⇒ x + 1 = y+2 = 3 + 3$

$⇒$ $\left\{\begin{matrix}x = 5& \\ y = 4& \end{matrix}\right.$

  Vậy `(x;y;z)=(1/5;-4/5;-9/5);(1/2; -1/2;-3/2);(5;4;3)`.