$\frac{x-1}{2}$ = $\frac{y-2}{3}$ = $\frac{z-3}{4}$ và x-2y+3z=-10 24/08/2021 Bởi Charlie $\frac{x-1}{2}$ = $\frac{y-2}{3}$ = $\frac{z-3}{4}$ và x-2y+3z=-10
`{x-1}/2={y-2}/3={z-3}/4={x-1}/2={2y-4}/6={3z-9}/12={x-1-2y+4+3z-9}/{2-6+12}={-16}/{8}=-2` $⇒\begin{cases}\dfrac{x-1}{2}=-2⇒x-1=-4⇒x=-3\\\dfrac{y-2}{3}=-2⇒y-2=-6⇒y=-4\\\dfrac{z-3}{4}=-2⇒z-3=-8⇒z=-5\end{cases}$ `Vậy` `(x;y;z)=(-3;-4;-5)` Bình luận
Giải thích các bước giải: Ta có : $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4}$ $\to \dfrac{x-1}{2} = \dfrac{2y-4}{6} = \dfrac{3z-9}{12} = \dfrac{x-1-(2y-4)+(3x-9)}{2-6+12} = \dfrac{(x-2y+3z)-6}{8} = \dfrac{-10-6}{8} =-2$ $\to \left\{ \begin{array}{l}\dfrac{x-1}{2} = -2\\\dfrac{y-2}{3} = -2\\\dfrac{z-3}{4} = =2\end{array} \right.$ $\to x = -3,y=-4,z=-5$ Vậy $(x,y,z) = (-3,-4,-5)$ Bình luận
`{x-1}/2={y-2}/3={z-3}/4={x-1}/2={2y-4}/6={3z-9}/12={x-1-2y+4+3z-9}/{2-6+12}={-16}/{8}=-2`
$⇒\begin{cases}\dfrac{x-1}{2}=-2⇒x-1=-4⇒x=-3\\\dfrac{y-2}{3}=-2⇒y-2=-6⇒y=-4\\\dfrac{z-3}{4}=-2⇒z-3=-8⇒z=-5\end{cases}$
`Vậy` `(x;y;z)=(-3;-4;-5)`
Giải thích các bước giải:
Ta có : $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4}$
$\to \dfrac{x-1}{2} = \dfrac{2y-4}{6} = \dfrac{3z-9}{12} = \dfrac{x-1-(2y-4)+(3x-9)}{2-6+12} = \dfrac{(x-2y+3z)-6}{8} = \dfrac{-10-6}{8} =-2$
$\to \left\{ \begin{array}{l}\dfrac{x-1}{2} = -2\\\dfrac{y-2}{3} = -2\\\dfrac{z-3}{4} = =2\end{array} \right.$
$\to x = -3,y=-4,z=-5$
Vậy $(x,y,z) = (-3,-4,-5)$