Question

giải hệ phương trình :

\(\dfrac{x-1}{5}\)= \(\dfrac{y-2}{3}\) = \(\dfrac{z-2}{2}\)

3x - 2y + z = 12

Answer 1

\(\left\{{}\begin{matrix}\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}\left(1\right)\\3x-2y+z=12\left(2\right)\end{matrix}\right.\)

* Xét phương trình (1) :

\(\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}\)= \(\dfrac{3x-3}{15}=\dfrac{2y-4}{6}=\dfrac{z-2}{2}\)

= \(\dfrac{\left(3x-3\right)-\left(2y-4\right)+\left(z-2\right)}{15-6+2}\)( Áp dụng dãy tỉ số bằng nhau )

\(\Rightarrow\)\(\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}\)= \(\dfrac{\left(3x-2y+z\right)-1}{11}\) = \(\dfrac{12-1}{11}\) ( vì 3x-2y+z=12)

\(\Rightarrow\) \(\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}\)=1

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=5\\y-2=3\\z-2=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=5\\z=4\end{matrix}\right.\)