Ta có:
\(\dfrac{x}{5}=\dfrac{y}{-7};\dfrac{y}{4}=\dfrac{z}{15}\)
\(\Rightarrow\dfrac{x}{-20}=\dfrac{y}{28}=\dfrac{z}{105}\)
\(\Rightarrow\dfrac{x}{-20}=\dfrac{y}{28}=\dfrac{z}{105}=\dfrac{x}{-20}=\dfrac{3y}{84}=\dfrac{4z}{420}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{-20}=\dfrac{3y}{84}=\dfrac{4z}{420}=\dfrac{x+3y-4z}{-20+84-420}=\dfrac{18}{-356}=-\dfrac{9}{178}\)
\(\Leftrightarrow\dfrac{x}{-20}=-\dfrac{9}{178}\Rightarrow x=\dfrac{90}{89}\)
\(\Leftrightarrow\dfrac{y}{28}=-\dfrac{9}{178}\Rightarrow y=-\dfrac{126}{89}\)
\(\Leftrightarrow\dfrac{z}{105}=-\dfrac{9}{178}\Rightarrow z=-\dfrac{945}{178}\)
Vậy ...
Lời giải:
Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{-7}\\\dfrac{y}{4}=\dfrac{z}{15}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{y}{-28}\\\dfrac{y}{-28}=\dfrac{z}{-105}\end{matrix}\right.\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{-28}=\dfrac{z}{-106}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{20}=\dfrac{y}{-28}=\dfrac{z}{-105}=\dfrac{3y}{84}=\dfrac{4z}{420}=\dfrac{x+3y-4z}{20+84-420}=\dfrac{18}{-316}=-\dfrac{9}{158}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=20.\dfrac{-9}{158}\\y=-28.\left(\dfrac{-9}{158}\right)\\z=-105.\left(-\dfrac{9}{158}\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{90}{79}\\y=\dfrac{126}{79}\\z=\dfrac{945}{158}\end{matrix}\right.\)
Vậy ...
