Ta có: \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
\(\Leftrightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
mà 2x+3y-z=95
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-3+3y-6-z+3}{4+9-4}=\dfrac{95-6}{9}=\dfrac{89}{9}\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{2x-2}{4}=\dfrac{89}{9}\\\dfrac{3y-6}{9}=\dfrac{89}{9}\\\dfrac{z-3}{4}=\dfrac{89}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-2=\dfrac{356}{9}\\3y-6=89\\z-3=\dfrac{356}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{374}{9}\\3y=95\\z=\dfrac{383}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{187}{9}\\y=\dfrac{95}{3}\\z=\dfrac{383}{9}\end{matrix}\right.\)
Vậy: (x,y,z)=\(\left(\dfrac{187}{9};\dfrac{95}{3};\dfrac{383}{9}\right)\)
