\(a.\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
\(=>\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-2y+4+3z-9}{2-6+12}=\dfrac{x-2y+3z-6}{8}=\dfrac{14-6}{8}=1\)
\(=>\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.=>\left\{{}\begin{matrix}x-1=2\\y-2=3\\z-3=4\end{matrix}\right.=>\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
\(b,\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=>\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=\dfrac{2x+3y-z-5}{9}=\dfrac{50-5}{9}=5\)
\(=>\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.=>\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.=>\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)
