a) Ta có:\(\left\{{}\begin{matrix}4x=5y\Rightarrow x=\dfrac{5y}{4}\\3y=7z\Rightarrow z=\dfrac{3y}{7}\end{matrix}\right.\)
\(x-3y+2z=-25\Rightarrow\dfrac{5y}{4}-3y+2.\dfrac{3y}{7}=-25\Rightarrow y=28\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}=35\\z=\dfrac{3y}{7}=12\end{matrix}\right.\)
b) \(\dfrac{x-2}{3}=\dfrac{y-3}{5}=\dfrac{z+1}{4}\Rightarrow20x-40=12y-36=15z+15\Rightarrow\left\{{}\begin{matrix}20x-40=15z+15\\12y-36=15z+15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3z+11}{4}\\y=\dfrac{15z+51}{12}\end{matrix}\right.\)
\(2x+3y-4z=22\Rightarrow2.\dfrac{3x+11}{4}+3.\dfrac{15z+51}{12}-4z=22\Rightarrow z=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3z+11}{4}=5\\y=\dfrac{15z+51}{12}=8\end{matrix}\right.\)
b: Ta có: \(\dfrac{x-2}{3}=\dfrac{y-3}{5}=\dfrac{z+1}{4}\)
\(\Leftrightarrow\dfrac{2x-4}{6}=\dfrac{3y-9}{15}=\dfrac{4z+4}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-4}{6}=\dfrac{3y-9}{15}=\dfrac{4z+4}{16}=\dfrac{2x+3y-4z-4-9-4}{6+15-16}=\dfrac{22-17}{5}=1\)
Do đó: \(\left\{{}\begin{matrix}2x-4=6\\3y-9=15\\4z+4=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=8\\z=3\end{matrix}\right.\)
a:Ta có: 4x=5y
nên \(\dfrac{x}{5}=\dfrac{y}{4}\)
hay \(\dfrac{x}{35}=\dfrac{y}{28}\left(1\right)\)
Ta có: 3y=7z
nên \(\dfrac{y}{7}=\dfrac{z}{3}\)
hay \(\dfrac{y}{28}=\dfrac{z}{12}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\)suy ra \(\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{12}\)
hay \(\dfrac{x}{35}=\dfrac{3y}{84}=\dfrac{2z}{24}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{35}=\dfrac{3y}{84}=\dfrac{2z}{24}=\dfrac{x-3y+2z}{35-84+24}=\dfrac{-25}{25}=1\)
Do đó: x=35; y=28; z=12
