Coi đề lại câu a
b,
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\\ \dfrac{x-1}{2}=\dfrac{2\left(y-2\right)}{2\cdot3}=\dfrac{3\cdot\left(z-3\right)}{3\cdot4}\\ \dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-\left(2y-4\right)+3z-9}{2-6+12}=\dfrac{x-1-2y+4+3z-9}{8}=\dfrac{\left(x-2y+3z\right)+\left(4-1-9\right)}{8}=\dfrac{14+\left(-6\right)}{8}=\dfrac{8}{8}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\Rightarrow x-1=2\Rightarrow x=3\\\dfrac{2y-4}{6}=1\Rightarrow2y-4=6\Rightarrow2y=10\Rightarrow y=5\\\dfrac{3z-9}{12}=1\Rightarrow3z-9=12\Rightarrow3z=21\Rightarrow z=7\end{matrix}\right.\)
Vậy x = 3; y = 5; z = 7
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
\(\Rightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}\)
\(=\dfrac{14-6}{14-6}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
