Question

[ỵyjfdfj]

\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{7}\) và \(2x+3y-z=372\)

Answer 1

[Minh Hiếu]

\(\dfrac{x}{3}=\dfrac{y}{4}\)⇒\(\dfrac{x}{15}=\dfrac{y}{20}\)

\(\dfrac{y}{5}=\dfrac{z}{7}\)⇒\(\dfrac{y}{20}=\dfrac{z}{28}\)

⇒\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)⇒\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)

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

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{372}{62}=6\)

⇒\(\left\{{}\begin{matrix}x=6.15=90\\y=6.20=120\\z=6.28=168\end{matrix}\right.\)

Answer 2

[Lấp La Lấp Lánh]

\(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{372}{62}=6\)

\(\Rightarrow\left\{{}\begin{matrix}x=6.15=90\\y=6.20=120\\z=6.28=168\end{matrix}\right.\)