Question

Tìm các số hữu tỉ x, y, z:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) và \(x-2y+3z=-10\)

Answer 1

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)\(\Leftrightarrow\dfrac{x-1}{2}=\dfrac{2\left(y-2\right)}{6}=\dfrac{3\left(z-3\right)}{12}\)

Suy ra: \(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{\left(x-1\right)}{2}=\dfrac{\left(2y-4\right)}{6}=\dfrac{\left(3z-9\right)}{12}=\dfrac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{2-6+12}\)

\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}=\dfrac{\left(x-2y+3z\right)+\left(4-1-9\right)}{8}\)

\(=\dfrac{-10-6}{8}=\dfrac{-16}{8}=2\)

Suy ra:

\(\left\{{}\begin{matrix}x=2.2+1=5\\y=2.3+2=8\\z=2.4+3=11\end{matrix}\right.\)