Đáp án: $(x,y,z)=(2,3,4)$
Giải thích các bước giải:
$\dfrac{1+2x}{5}=\dfrac{3+y}{6}=\dfrac{2z-1}{7}$
$\rightarrow\dfrac{1+2x}{5}=\dfrac{3+y}{6}=\dfrac{2z-1}{7}= \dfrac{2+4x}{10}=\dfrac{6+2y}{12}=\dfrac{2z-1}{7}=\dfrac{2+4x-(6+2y)-(2z-1)}{10-12-7}$
$\rightarrow \dfrac{1+2x}{5}=\dfrac{3+y}{6}=\dfrac{2z-1}{7}= \dfrac{2(2x-y-z)-3}{-9}=\dfrac{2.(-3)-3}{-9}=1$
$\rightarrow \begin{cases}1+2x=5\rightarrow x=2\\3+y=6\rightarrow y=3\\2z-1=7\rightarrow z=4\end{cases}$
