Question

Tìm x,y,z biết:

\(\dfrac{x}{3}\)=\(\dfrac{3y}{2}\) ; \(\dfrac{4y}{3}\)=\(\dfrac{z}{5}\) và 2x+3y-z=-7

Answer 1

Ta có x/3=3y/2

nên \(\dfrac{x}{3}=\dfrac{y}{\dfrac{2}{3}}\)

\(\Leftrightarrow\dfrac{x}{3}\cdot\dfrac{1}{6}=\dfrac{y}{\dfrac{2}{3}}\cdot\dfrac{1}{6}\)

=>x/18=y/4

=>x/54=y/12

\(\dfrac{4y}{3}=\dfrac{z}{5}\)

nên \(\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{5}\)

\(\Leftrightarrow\dfrac{y}{\dfrac{3}{4}}\cdot\dfrac{1}{60}=\dfrac{z}{5}\cdot\dfrac{1}{60}\)

=>y/45=z/300

=>y/3=z/20

=>y/12=z/80

=>x/54=y/12=z/80

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{54}=\dfrac{y}{12}=\dfrac{z}{80}=\dfrac{2x+3y-z}{2\cdot54+3\cdot12-80}=\dfrac{-7}{64}\)

Do đó: \(\left\{{}\begin{matrix}x=-\dfrac{189}{32}\\y=-\dfrac{21}{16}\\z=-\dfrac{35}{4}\end{matrix}\right.\)