Question

Tìm x,y,z biết \(\frac{x-1}{2}=\)\(\frac{y-2}{3}\)= \(\frac{z-3}{4}\)và x-2y+3z = 14

Answer 1

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

\(\Rightarrow\frac{x-1}{2}=\frac{2\left(y-2\right)}{2\cdot3}=\frac{3\left(z-3\right)}{3\cdot4}\)

\(\Rightarrow\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)

theo tính chất của dãy tỉ số bằng nhau ta có :

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

\(=\frac{\left(x-2y+3z\right)+\left(-1+4-9\right)}{8}=\frac{14+\left(-6\right)}{8}=\frac{8}{8}=1\)

\(\Rightarrow\hept{\begin{cases}x=1\cdot2+1=3\\y=1\cdot3+2=5\\z=1\cdot4+3=7\end{cases}}\)

vậy_